Wave Motion and String Waves

Chapter # 15

1.

Wave Motion and Waves on A string

SOLVED EXAMPLES

A wave is propagating on a long stretched string along its length taken as the positive x-axis. The wave equation is given as x  t    T   e

Sol.

2

y = y0 where y0 = 4 mm, T = 1.0 s and  = 4 cm. (a) Find the velocity of the wave. (b) Find the function f(t) giving the displacement of the particle at x = 0. (c) Find the function g(x) giving the shape of the string at t = 0. (d) Plot the shape g(x) of the string at t = 0. (e) Plot the shape of the string at t = 5 s. (a) the wave equation may be written as 

1 x  t   /T  T2 

2

y = y0 e comparing with the general equation y = f(t – x/v), we see that

(b)

4 cm  = 1.0 s = 4 cm/s. T putting x = 0 in the given equation, .............(i)

(c)

f(t) = y0 e ( t / T ) putting t = 0 in the given equation g(x) = y0 e ( x /  )

.............(ii)

v=

2

2

(d)

(e) 2.

Sol.

Consider the wave y = (5 mm) sin (1 cm–1) x – (60 s–1) t]. Find (a) the amplitude (b) the wave number, (c) the wavelength, (d) the frequency, (e) the time period and (f) the wave velocity. x  Comparing the given equation with equation A sin    t  , v  we find (a) amplitude A = 5 mm (b) wave number k = 1 cm–1

(c)

wavelength

=

2 = 2 cm k

(d)

frequency

v=

 60 = Hz 2 2

(e)

time period

T=

(f)

wave velocity

 1 = s 30 v u = v = 60 cm/s.

3.

Figure shows a string of linear mass density 1.0 g/cm on which a wave pulse is travelling. Find the time taken by the pulse in travelling through a distance of 50 cm on the string. Take g = 10 m/s2.

Sol.

The tension in the string is F = mg = 10 N. The mass per unit length is µ = 1.0 g/cm = 0.1 kg/m. The wave manishkumarphysics.in

Page # 1

Chapter # 15


velocity is, therefore, v = F / µ =

10 N 0.1 kg / m = 10 m/s. The time taken by the pulse in travelling through 50

cm is, therefore, 0.05 s. 4.

Sol.

The average power transmitted through a given point on a string supporting a sine wave is 0.20 W when the amplitude of the wave is 2.0 mm. What power will be transmiited through this point if the amplitude is increased to 3.0 mm. Other things remaining the same, the power transmitted is proportional to the square of the amplitude.

P2 A 22  P1 A12

Thus,

P2 9 0.20 W = 4 = 2.25 P2 = 2.25 × 0.20 W = 0.45 W.

or, or, 5.

Two waves are simultaneously passing through a string. The equations of the waves are given by y1 = A1 sin k (x – vt) and y2 = A2 sin k (x – vt + 1)  where the wave number k = 6.28 cm –1 and x 0 = 1.50 cm. The amplitudes are A1 = 5.0 mm and A2 = 4.0 mm. Find the phase difference between the waves and the amplitude of the resulting wave.

Sol.

The phase of the first wave is k (x – vt) and of the second is k (x – vt + x0). the phase difference is, therefore  = kx0 = (6.28 cm–1) (1.50 cm) = 2 × 1.5 = 3. The waves satisfy the condition of destructive interference. The amplitude of the resulting wave is given by | A1 – A2 | = 5.0 mm – 4.0 mm = 1.0 mm.

6.

Two travelling waves of equal amplitudes and equal frequencies move in opposite direction along a string. They interfere to produce a standing wave having the equation. y = A cos kx sin t in which A = 1.0 mm, k = 1.57 cm–1 and  = 78.5 s–1. (a) Find the velocity of the component travelling waves. (b) Find the node closest to the origin in the region x > 0. (c) Find the antinode closest to the origin in the region x > 0. (d) Find the amplitude of the particle at x = 2.33 cm.

Sol.

(a)

The standing wave is formed by the superposition of the waves y1 =

A sin (t – kx) 2

and

A sin (t + kx). 2 The wave velocity (magnitude) of either of the waves is y2 =

78.5 s 1  v= = = 50 cm/s. k 1.57 cm 1

(b)

For a node, cos kx = 0. The smallest positive x satisfying this relation is given by kx = /2

3.14  = = 1 cm 2  1.57 cm 1 2k (c) For an antinode, |cos kx| = 1. The smallest positive x satisfying this relation is given by kx = 

or,

or, (d)

x=

 = 2 cm k The amplitude of vibration of the particle at x is given by | A cos kx |. For the given point, x=

kx = (1.57 cm–1) (2.33 cm) =

7  =+ . 6 6

Thus, the amplitude will be manishkumarphysics.in

Page # 2

Chapter # 15

Wave Motion and Waves on A string (1.0 mm) | cos ( + /6) | =

3 mm = 0.86 mm. 3

7.

A 50 cm long wire of mass 20 g supports a mass of 1.6 kg as shown in figure. Find the fundamental frequency of the portion of the string between the wall and the pulley. Take g = 10 m/s2.

Sol.

The tension in the string is

F = (1.6 kg) (10 m/s2) = 16 N.

The linear mass density is

20 g µ = 50 cm 0.04 kg/m.

The fundamental frequency is

1 v0 = 2 L

F/µ =

1 2  (0.4m)

16 N 0.04 kg / m = 25 Hz.

8.

In a sonometer experiment, resonance is obtained when the experimental wire has a length of 21 cm between the bridges and the vibrations are excited by a tuning fork of frequency 256 Hz. If a tuning fork of frequency 384 Hz is used, what should be the length of the experimental wire to get the resonance ?

Sol.

By the law of length, or,

l2 =

v1 v2

l1 =

l1 v1 = l2 v2 256 × 21 cm = 14 cm. 384

QUESTIONS

FOR

SHORT

ANSWER

1.

You are walking along a seashore and a mild wind is blowing. Is the motion of air a wave motion?

2.

The radio and TV programmes, telecast at the studio, reach our antenna by wave motion. Is it a mechanical wave or nonmechanical?

3.

A wave is represented by an equation y = c1 sin (c2x + c3t). In which direction is the wave going? Assume that c1, c2 and c3 are all positive.

4.

Show that the particle speed can never be equal to the wave speed in a sine wave if the amplitude is less than wavelength divided by 2.

5.

Two wave pulse identical in shape but inverted with respect to each other are produced at the two ends of a stretched string. At an instant when the pulses reach the middle, the string becomes completely straight. What happens to the energy of the two pulses? [HCV_Chp 15_ Q.S.A_5]

6.

Show that for a wave travelling on a string

y max v max = v max amax , where the symbols have usual meanings. Can we use componedo and dividendo taught in algebra to write v max  amax y max  v max = ? v max  a max y max  v max 7.

What is the smallest positive phase constant which is equivalent to 7.5?

8.

A string clamped at both ends vibrates in its fundamental mode. Is there any position (except the ends) on the string which can be touched without disturbing the motion? What if the string vibrates in its first overtone ? manishkumarphysics.in

Page # 3

Chapter # 15


Objective - I

1.

A sine wave is travelling in a medium. The minimum distance between the two particles, always having same speed, is HCV_Ch-15_Obj I_1 fdlh ek/;e esa ,d T;k rjax xfreku gSA ges'kk leku pky ds nks d.kksa ds chp dh U;wure nwjh gS (A) /4 (B) /3 (C*) /2 (D) 

2.

A sine wave is travelling in a medium. A particular particle has zero displacement at a certain instant. The particle closest to it having zero displacement is at a distance HCV_Ch-15_Obj I_2

fdlh ek/;e esa T;k rjax xfreku gSA fuf'pr {k.k ij ,d fo'ks"k d.k dk foLFkkiu 'kwU; gSA bl d.k ls ml d.k dh U;wure nwjh ftldk Hkh foLFkkiu 'kwU; gSA (A) /4

(B) /3

(C*) /2

(D) 

3.

Which of the following equations represents a wave travelling along Y-axis? Y-v{k ds vuqfn'k xfreku rjax dk lehdj.k gS ? (A*) x = A sin (ky – t) (B) y = A sin (kx – t) (C) y = A sin ky cos t (D) y = A cos ky sin t.

4.

The equation y = A sin2 (kx – t) represents a wave motion with HCV_Ch-15_Obj I_4 (A) amplitude A, frequency /2 (B*) amplitude A/2, frequency / (C) amplitude 2A, frequency /4 (D) does not represent a wave motion

5.

Which of the following is a mechanical wave ? (A) radio waves (B) x-rays fuEu esa ;kaf=kd rjax gS ? (A) jsfM;ks rjaxs (B) x-fdj.k

HCV_Ch-15_Obj I_3

HCV_Ch-15_Obj I_5

(C) light waves

(D*) sound waves

(C) izdk'k

(D*) /ofu

rjax

rjaxs

6.

A cork floating in a calm pond executes simple harmonic motion of frequency v when a wave generated by a boat passes by it. The frequency of the wave is HCV_Ch-15_Obj I_6 'kkUr ty esa ,d dkdZ dk VqdM+k v vko`fr ls ljy vkorZ xfr dj jgk gSA tc ,d uko ls mRiUu rjaxs blls xqtjrh gS rks rjax dh vko`fRr gksxh (A*) v (B) v/2 (C) 1/4 (D) 4

7.

Two strings A and B, made of same material, are stretched by same tension. The radius of string A is double of the radius of B. A transverse wave travels on A with speed vA and on B with speed vB. The ratio vA/vB is HCV_Ch-15_Obj I_7

nks jfLl;k¡ A o B, ,d gh inkFkZ dh cuh gqbZ gS rFkk nksuksa esa leku ruko gSA A dh f=kT;k B dh nqxuh gSA ,d vuqiLz Fk rjax A ij VA pkyls rFkk B ij VB pky ls xfreku gSA vA / vB dk vuqikr gS (A*) 1/2 8.

(B) 2

(C) 1/4

(D) 4

Both the strings, show in figure, are made of same material and have same cross-section. The pulleys are light. The wave speed of a transverse wave in the string AB is v1 and in CD it is v2. The v1/v2 is HCV_Ch-15_Obj I_8

fp=k esa fn[kkbZ xbZ jfLl;k¡ ,d gh inkFkZ rFkk ,d gh leku ruko ij gSA f?kjuh;k¡ gYdh gSAa jLlh AB esa vuqiLz Fk rjaxksa dk V1 rFkk C D esa osx v2 gS rks v1 / v2 gksxk

(A) 1

(B) 2

(C)

2

manishkumarphysics.in

(D*) 1/ 2

Page # 4

Chapter # 15 Wave Motion and Waves on A string 9. Velocity of sound in air is 332 m/s. Its velocity in vacuum will be HCV_Ch-15_Obj I_9 ok;q esa /ofu dk osx 332 eh@ls gSA bldk fuokZr esa osx gksxk (A) > 332 m/s (B) = 332 m/s (C) < 332 m/s (D*) meaningless 10.

A wave pulse, travelling on a two piece string, gets partially reflected and partially transmitted at the junction. The reflected wave is inverted in shape as compared to the incident one. If the incident wave has wavelength  and the transmitted wave , HCV_Ch-15_Obj I_10

,d rjax Lian nks VqdM+kas ls cuh gqbZ Mksjh esa xeu dj jgk gS] ;g laf/k ij vkaf'kd ijkofrZr rFkk vkaf'kd ikjxfer gksrh gSA ijkofrZr rjax dh vkÑfr] vkifrr dh rqyuk esa mRØfer gks tkrh gSA ;fn vkifrr rjax dh rjaxnS/;Z  gS rFkk ikjxfer rjax dh , gS (A)  >  (B)  =  (C*)  <  (D) nothing can be said about the relation of  and . 11.

Two waves represented by y = a sin (t – kt) and y = a cos(t – kx) are superposed. The resultant wave will have an amplitude HCV_Ch-15_Obj I_11 y = a sin (t – kt) vkSj y = a cos(t – kx) }kjk iznf'kZr rjax v/;kjksfir gSA ifj.kkeh rjax dk vk;ke gksxk (A) a

(B*)

2a

(C) 2a

(D) 0

12.

Two wires A and B, having identical geometrical construction, are stretched from their natural length by small but equal amount. The Young’s modulus of the wires are YA > YB and A > B. A transverse signal started at one end takes a time t1 to reach the other end for A and t2 for B. HCV_Ch-15_Obj I_12 nks rkj A o B ftudh T;kfefr vkÑfr ,d leku gSA nksuksa dks mudh okLrfod yEckbZ ls leku ysfdu okLrfod yEckbZ ls leku ysfdu NksVh ek=kk esa [khapk x;k gSA rkjksa dk ;ax izR;kLFkrk xq.kkad YA > YB rFkk ?kuRo A > B gSA ,d fljs ij mRiUu ,d fljs ij mRiUu ladsr nwljs fljs ij igq¡pus esa A ds fy, t1 le; rFkk B ds fy, t2 le; ysrk gS rks (A) t1 < t2 (B) t1 = t2 (C) t1 > t2 (D*) the information is insufficient to find the relation between t1 and t2. t1 o t2 esa lac/a k Kkr djusds fy, lwpuk vi;kZIrA

13.

Consider two waves passing through the same string. Principle of superposition for displacement says that the net displacement of a particle on the string is sum of the displacements produced by the two waves individually. Suppose we state similar principles for the net velocity of the particle and the net kinetic energy of the particle. Such a principle will be valid for HCV_Ch-15_Obj I_13 (A) both the velocity and the kinetic energy (B*) the velocity but not for the kinetic energy (C) the kinetic energy but not for the velocity (D) neither the velocity nor the kinetic energy

nks rjaxksa dks ,d gh jLlh ls xqtjrh gqbZ ekurs gq, foLFkkiu ds fy, v/;kjksi.k ds fl)kUr ds vuqlkj jLlh ds fdlh d.k dk dqy foLFkkiu nksuksa rjaxksa }kjk mRiUu vyx&vyx foLFkkiuksa ds ;ksx ds cjkcj gksrk gSA ;fn ge ;gh fl)kUr d.kksads dqy osx vkSj dqy xfrt ÅtkZ ds fy, fy[ksa rks bl izdkj dk fl)kUr ykxw gksxk (A) osx vkSj xfrt mtkZ nksuksa ds fy, (B*) osx ds fy, ysfdu xfrt mtkZ ds fy, ugha (C) xfrt mtkZ ds fy,] ysfdu osx ds fy, ugha (D) osx vkSj xfrt mtkZ nksuksa ds fy, ugha 14.

Two wave pulses travel in opposite directions on a string and approach each other. The shape of the one pulse in inverted with respect to the other. HCV_Ch-15_Obj I_14 (A) the pulses will collide with each other and vanish after collision (B) the pulses will reflect from each other i.e., the pulse going towards right will finally move towards left and vice versa. (C) the pulses will pass through each other but their shapes will be modified (D*) the pulses will pass through each other without any change in their shape

nks rjax Lian ,d jLlhij foifjr fn'kkvksa esa xfreku gS rFkk ,d nwljs dh rjQ vkrh gSA ,d Lian dk vkdkj nwljs ds lkis{k mYVk gSA HCV_Ch-15_Obj I_14 (A) Lian ,d nwljs ls Vdjk;saxs rFkk bl VDdj ds ckn [kRe gks tk;saxsA (B) Lian ,d nwljs ls ijkofrZr gksx a s vFkkZr~ nk;ha rjQ tkus okyh Lian vUr esa ck;ha rjQ tk;sxk rFkk blh izdkj ck;ha cká nk;ha rjQ (C) Lian ,d nwljs ls xqtj tk;saxs ysfdu mudh vkÑfr cnysxhA (D*) Lian ,d nwljs ls xqtj tk;saxs ysfdu mudh vkÑfr ugha cnysxhA


Page # 5

Chapter # 15 Wave Motion and Waves on A string 15. Two periodic waves of amplitudes A1 and A2 pass through a region. If A1 > A2 , the difference in the maximum and minimum resultant amplitude possible is : HCV_Ch-15_Obj I_15 A1 rFkk A2 vk;ke okyh nks vkorhZ rjaxsa fdlh {ks=k ls xqtjrh gSA ;fn A1 > A2 gS] rks lEHko vf/kdre rFkk U;wure

ifj.kkeh vk;ke dk vUrj gS & (A) 2A1

16.

(B*) 2A2

(C) A1 + A2

(D) A1 – A2

Two waves of equal amplitude A, and equal frequency travels in the same direction in a medium. The amplitude of the resultant wave is HCV_Ch-15_Obj 1_16

leku vk;ke rfkk leku vko`fÙk dh nks rjaxs fdlh ek/;e esa,d gh fn'kk esa ,d gh fn'kk esa xeu dj jgh gSA ifj.kkeh rjax dk vk;ke gS & (A) 0 (0 rFkk 2A ds 17.

19.

(C) 2A

(D*) between 0 and 2A.

Two sine waves travel in the same direction in a medium. The amplitude of each wave is A and the phase difference between the two waves is 120°. The resultant amplitude will be fdlh ek/;e esa nks T;koØh; rjaxsa ,d gh fn'kk esa xeu dj jgh gSA izR;sd rjax dk vk;ke A gS rFkk nksuksa rjaxksa esa dykUrj 120° gSA ifj.kkeh vk;ke gksxk & (A*) A

18.

(B) A

chp)

(B) 2A

(C) 4A

The fundamental frequency of a string is proportional to (A*) inverse of its length (B) the diameter (C) the tension

(D)

2A

(D) the density

A tuning fork of frequency 480 Hz is used to vibrate a sonometer wire having natural frequency 240 Hz. The wire will vibrate with a frequency of HCV_Ch-15_Obj I_19 240 gV~Zt izkd`frd vko`fÙk okys Lojekih ds rkj dks dfEir djus ds fy;s] 480 gV~Zt vko`fÙk dk Lofj=k f}Hkqt iz;qDr

fd;k tkrk gSA rkj fuEu vko`fÙk ls dEiUu djsxk & (A) 240 Hz (A) 240 gV~Zt 20.

(B*) 480 Hz (B*) 480 gV~Zt

(C) 720 Hz (C) 720 gV~Zt

(D) will not vibrate (D) dEiUu ugha gksxkA

A tuning fork of frequency 480 Hz is used to vibrate a sonometer wire having natural frequency 410 Hz. The wire will vibrate with a frequency HCV_Ch-15_Obj I_20 410 gV~t Z ewy dh vko`fÙk okys Lojekih ds rkj dks dfEir djus ds fy;s] 480 gVZ~t dh vko`fÙk dk Lofj=k f}Hkqt iz;D q r

fd;k tkrk gSA rkj ftl vko`fÙk ls dEiUu djsxk] og gS & (A) 410 Hz (A) 410 gV~Zt 21.

(B*) 480 Hz (B*) 480 gV~Zt

(C) 820 Hz (C) 820 gV~Zt

(D) 960 Hz (D) 960 gV~Zt

A sonometer wire of length l vibrate in fundamental mode when excited by a tuning fork of frequency 416 Hz. If the length is doubled keeping other things same, the string will tc 416 gV~Zt vko`fÙk ds Lofj=k f}Hkqt }kjk mÙksftr fd;k tkrk gSA rks Lojekih dk l yEckbZ dk rkj bldh ewy

fo/kk esa dEiUu djrk gSA ;fn vU; jkf'k;ksa dks leku j[krs gq,] ;fn yEckbZ nqxuh dj nh tk;s] rks Mksjh & (A*) vibrate with a frequency of 416 Hz (C) vibrate with a frequency of 832 Hz

(A*) 416 gV~Zt vko`fÙk ls dEiUu djsxhA (C) 832 gV~Zt 22.

vko`fÙk ls dEiUu djsxhA

(B) vibrate with a frequency of 208 Hz (D) stop vibrating (B) 208 gV~Zt vko`fÙk ls dEiUu djsxhA (D) dEiUu djuk cUn dj nsxhA

A sonometer wire supports a 4 kg and load and vibrates in fundamental mode with a tuning fork of frequency 416 Hz. The length of the wire between the bridges is now doubled. In order to maintain fundamental mode, the load should be changed to ,d Lojekih ds rkj ls 4 fdxzk- Hkkj yVdk;k x;k gS rFkk ;g 416 gV~Zt vko`fÙk ds Lofj=k ds lkFk ewy fo|k esa dEiUu

djrk gSA vc lsrv q ksa ds e/; rkj dh yEckbZ nqxuh dj nh tkrh gSA ewy fo|k dks lqfuf'pr djus ds fy;s] Hkkj ifjofrZr djds djuk gksxk & (A) 1 kg

(B) 2 kg

(C) 8 kg


(D*) 16 kg.

Page # 6

Chapter # 15

1.


Objective - II

A mechanical wave propagates in a medium along the X-axis. The particles of the medium HCV_Ch-15_Obj II_1

(A) must move on the X-axis (C*) may move on the X-axis

(B) must move on the Y-axis (D*) may move on the Y-axis

X-v{k ds vuqfn'k ek/;e esa ,d ;kaf=kd (A) X-v{k ij xfr djuk gh pkfg, (C*) X-v{k ij xfr dj ldrk gS

rjax xfreku gSA ek/;e ds d.k (B) Y-v{k ij xfr djuk gh pkfg, (D*) Y-v{k ij xfr dj ldrk gS

2.

A transverse wave travels along the Z-axis. The particles of the medium must move (A) along the Z-axis (B) along the X-axis (C) along the Y-axis (D*) in the X-Y plane ,d vuqiLz Fk rjax Z-v{k ds vuqfn'k ,d ;kaf=kd rjax xeu dj jgh gSA ek/;e ds d.k& (A) Z-v{k ds vuqfn'k (B) X-v{k ds vuqfn'k (C) Y-v{k ds vuqfn'k (D*) X-Y ry esa

3.

Longitudinal waves cannot (A) have a unique wavelength (C) have a unique wave velocity

(B) transmit energy (D*) be polarized

vuqnS/;Z rjaxs esa ugh gksrk gS& (A) vf}rh; rjaxnS/;Z (C) vf}rh; rjax osx

(B) ÅtkZ LFkkukarj.k (D*) /kzfq or

A wave going in a solid (A) must be longitudinal (C) must be transverse

(B*) may be longitudinal (D*) may be transverse

Bksl esa xeu dj jgh rjax gksrh gS& (A) fuf'pr :i ls vuqnS/;Z (C) fuf'pr :i ls vuqiLz Fk gksrh gSA

(B*) vuqnS/;Z (D*) vuqiLz Fk

A wave moving in a gas (A*) must be longitudinal (C) must be transverse

(B) may be longitudinal (D) may be transverse

xSl esa xeu dj jgh rjax gksrh gS& (A*) fuf'pr :i ls vuqnS/;Z (C) fufp;r :i ls vuqizLFk gksrh gSA

(B) vuqnS/;Z (D) vuqiLz Fk

4.

5.

6.

HCV_Ch-15_Obj II_3

gks ldrh gSA gks ldrh gSA HCV_Ch-15_Obj II_5

gks ldrh gSA gks ldrh gSA

Two particles A and B have a phase difference of  when a sine wave passes through the region HCV_Ch-15_Obj II_6

(A) (B*) (C) (D*)

A oscillates at half the frequency of B A and B move in opposite directions A and B must be separated by half of the wavelength the displacements at A and B have equal magnitudes tc fdlh {ks=k ls ,d TokoØh; rjax xqtjrh gS] nks d.kks A o B ds e/; (A) A , B dh vk/kh vko`fr ls nksyu djsxkA (B*) A rFkk B foijhr fn'kk esa xfr djsxsaA (C) A o B ds e/; v/kZ rjaxnS/;Z dk vUrjky gksxkA (D*) A rFkk B ij foLFkkiu dk ifjek.k ,d leku gksxkA 7.

dykUrj  gS&

A wave is represented by the equation y = (0.001 mm) sin [(50 s–1) t + (2.0 m–1) x] HCV_Ch-15_Obj II_7 (A) the wave velocity = 100 m/s (B) the wavelength = 2.0 m (C*) the frequency = 25/ Hz. (D*) the amplitude = 0.001 mm. ,d rjax fuEu lehdj.k }kjk O;Dr dh tkrh gS% y = (0.001 mm) sin [(50 s–1) t + (2.0 m–1) x] (A) rjax osx = 100 m/s (B) rjaxnS/;Z = 2.0 m (C*) vko`fr = 25/ gVZ~t (D*) vk;ke = 0.001 mm.


Page # 7

Chapter # 15 Wave Motion and Waves on A string 8. A standing wave is produced on a string clamped at one end and free at the other. The length of the string HCV_Ch-15_Obj II_8

(A*) must be an integral multiple of /4 (C) must be an integral multiple of 

(B) must be an integral multiple of /2 (D) may be an integral multiple of /2

,d fljs ij dlh gqbZ rFkk nwljs fljs ij eqDr Mksjh esa vizxkeh rjax mRiUu gksrh gSA Mksjh dh yEckbZ& (A*) fuf'pr :i ls /4 dh iw.kZ xq.kt gksxhA (B) fuf'pr :i ls /2 dh iw.kZ xq.kt gksxhA (C) fuf'pr :i ls  dh iw.kZ xq.kt gksxhA (D) /2 dh iw.kZ xq.kt gks ldrh gSA 9.

Mark out the correct options. HCV_Ch-15_Obj II_9 (A) the energy of any small part of a string remains constant in a travelling wave (B*) the energy of any small part of a string remains constant in a standing wave (C) the energies of all the small parts of equal length are equal in a travelling wave (D) the energies of all the small parts of equal length are equal in a standing wave.

lgh dFku dk p;u dhft;s& (A) rjax ds xeku esa Mksjh ds fdlh vYi Hkkx dh ÅtkZ fu;r jgrh gSA (B*) vizxkeh rjax esa Mksjh ds fdlh Hkh vYi Hkkx dh ÅtkZ fu;r jgrh gSA (C) rjax ds xeku esa leku yEckbZ ds leLr vYi Hkkxksa dh ÅtkZ,¡ ,d leku gksrh gSA (D) vizxkeh rjax esa leku yEckbZ ds leLr vYi Hkkxks dh ÅtkZ,¡ ,d leku gksrh gS 10.

In a stationary wave, (A) all the particles of the medium vibrate in phase (B) all the antinodes vibrate in phase (C*) the alternate antinodes vibrate in phase (D*) all the particles between consecutive nodes vibrate in phase

HCV_Ch-15_Obj II_10

fdlh vizxkeh rjax esa& (A) ek/;e ds leLr d.k leku dyk esa dEiUu djrs gSA (B) leLr izLian leku dyk esa dEiUu djrs gSA (C*) ,dkarj izLian leku dyk esa dEiUu djrs gSA (D*) nks Øekxr fuLianks ds chp okys leLr d.k leku dyk esa dEiUu djrs gSA

WORKED OUT EXAMPLES 1.

The displacement of a particle of a string carrying a travelling wave is given by y = (3.0 cm) sin 6.28 (0.50x – 50t), where x is in centimeter and t in second. Find (a) the amplitude, (b) the wavelength, (c) the frequency and (d) the speed of the wave.

Sol.

Comparing with the standard wave equation y = Asin (kx – t)

x t  = A sin 2     T

we see that, amplitude = A = 3.0 cm, wavelength =  =

1 cm = 2.0 cm, 0.50

1 = 50 Hz. T The speed of the wave is u = v = (50 s–1) (2.0 cm) = 100 cm/s.

and the frequency = v =

2.

Sol.

The equation for a wave travelling in x-direction on a string is y = (3.0 cm) sin (3.14 cm–1) x – (314 s–1) t]. (a) Find the maximum velocity of a particle of the string. (b) Find the acceleration of a particle at x = 6.0 cm at time t = 0.11 s. (a) The velocity of the particle at x at time t is y = (3.0 cm) (– 314 s–1) cos [(3.14 cm–1 ) x – (314 s–1)t] t = (– 9.4 m/s) cos [(3.14 cm–1) x – (314 s–1) t].

v=


Page # 8

Chapter # 15

(b)

Wave Motion and Waves on A string The maximum velocity of particle will be v = 9.4 m/s. The acceleration of the particle at x at time t is v = – (9.4 m/s) (314 s–1) sin [(3.14 cm–1) x – (314 s–1) t ] t = (2952 m/s2) sin [(3.14 cm–1) x – (314 s–1) t] The acceleration of the particle at x = 6.0 cm at time t = 0.11 s is a = – (2952 m/s2) sin [6 – 11] = 0.

a=

3.

A long string having a cross-sectional area 0.80 mm2 and density 12.5 g/cm3, is subjected to a tension of 64 N along the X-axis. One end of this string is attached to a vibrator moving in transverse direction at a frequency of 20 Hz. At t = 0, the source is at a maximum displacement y = 1.0 cm. (a) Find the speed of the wave travelling on the string. (b) Write the equation for the wave. (c) What is the displacement of the particle of the string at x = 50 cm at time t = 0.05 s? (d) What is the velocity of this particle at this instant ?

Sol.(a) The mass of 1 m long part of the string is m = (0.80 mm2) × (1 m) × (12.5 g /cm3) = (0.80 × 10–6 m3) × (1.25 × 103 kg/m3) = 0.01 kg. The linear mass density is µ = 0.01 kg/m. The wave speed is u = = (b)

F/µ

64 N 0.01 kg / m = 80 m/s.

The amplitude of the source is A = 1.0 cm and the frequency is v = 20 Hz. The angular frequency is = 2v = 40 s–1. Also at t = 0, the displacement is equal to its amplitude i.e., at t = 0, x = A. The equation of motion of the source is, therefore, y = (1.0 cm) cos [(40 s–1)t] ..........(i) The equation of the wave travelling on the string along the positive X-axis is obtained by replacing t with t – x/ v in equation (i). It is, therefore,

 x  1  y = (1.0 cm) cos ( 40s )  t  v    

(c)

   1   1 = (1.0 cm cos ( 40 s ) t   2 m  x  , ...........(ii)     The displacement of the particle at x = 50 cm at time t = 0.05s is by the equation (ii),   1  y = (1.0 cm) cos [(40  s–1)] (0.05 s) –  m  (0.5 ml)] 2 

(d)

1.0 cm   = (1.0 cm) cos 2   = = 0.71 cm 4 2   The velocity of the particle at position x at time t is, by equation (ii), y v =  t = – (1.0 cm) (40 s–1) sin

   1   1 ( 40 s ) t   m  x  . 2   

Putting the values of x and t,

40   v = – (40 cm/s) sin  2   = cm/s = 89 cm/s 4 2  4.

Sol.

The speed of a transverse wave, going on a wire having a length 50 cm and mass 5.0 g, is 80 m/s. The area of cross-section of the wire is 1.0 mm2 and its Young’s modulus is 16 × 1011 N/m2. Find the extension of the wire over its natural length. The linear mass density is µ=

5  10 3 kg 50  10 2 m

= 1.0 × 10–2 kg/m.


Page # 9

Chapter # 15


The wave speed is v =

F/µ .

Thus, the tension is v = F = µv2  m2 2 kg   × 6400 = 1.0  10 = 64 N. m  s2 The Young’s modulus is given by

F/ A . L / L The extension is, therefore, Y=

L =

FL AY ( 64 N) (0.50 m)

=

(1.0  10

6

m 2 )  (16  1011N / m 2 )

= 0.02 mm.

5.

A uniform rope of length 12 m and mass 6 kg hangs vertically from a rigid support. A block of mass 2 kg is attached to the free end of the rope. A transverse pulse of wavelength 0.06 m is produced at the lower end of the rope. What is the wavelength of the pulse when it reaches the top of the rope?

Sol.

As the rope is heavy, its tension will be different at different points. The tension at the free end will be (2 kg) g and that at the upper end it will be (8 kg) g.

We have,

v = v

or,

F / µ = v

..........(i) F/ = v µ The frequency of the wave pulse will be the same everywhere on the rope as it depends only on the frequency or,

of the source. The mass per unit length is also the same throughout the rope as it is uniform. Thus, by (i)

F 

is constant. Hence,

(2 kg)g 0.06 m

=

(8 kg)g 1

,

where 1 is the wavelength at the top of the rope. This gives 1 = 0.12 m. 6.

Sol.

Two waves passing through a region are represented by y = (1.0 m) sin [(3.14 cm–1) x – (157 s–1) t] and y = (1.5 cm) sin [(1.57 cm–1)x – (314 s–1) t]. Find the displacement of the particle at x = 4.5 cm at time t = 5.0 ms. According to the principle of superposition, each wave produces its disturbance independent of the other and the resultant disturbance is equal to the vector sum of the individual disturbance. The displacements of the particle at x = 4.5 cm at time t = 5.0 ms due to the two waves are, y1 = (1.0 cm) sin [(3.14 cm–1) (4.5 cm)] – (157 s–1) (5.0 × 10–3 s)]   = (1.0 cm) sin 4.5   4 

1.0 cm   = (1.0 cm) sin 4   = 4 2  and y2 = (1.5 cm) sin [(1.57 cm–1)(4.5 cm) – (314 s–1) (5.0 × 10–3 s)]


Page # 10

Chapter # 15


    = (1.5 cm) sin 2.25    = (1.5 cm ) sin 2   2 4  

1.5 cm  =– 2 4 The net displacement is = (1.5 cm) sin

0.5 cm y = y1 + y2 = 7.

Sol.

2

= – 0.35 cm.

The vibrations of a string fixed at both ends are described by the equation y = (5.00 mm) sin [(1.57 cm–1) x] sin [(314 s–1)t]. (a) What is the maximum displacement of the particle at x = 5.66 cm? (b) What are the wavelengths and the wave speeds of the two transverse waves that combine to give the above vibration? (c) What is the velocity of the particle at x = 5.66 cm at time t = 2.00 s? (d) if the length of the string is 10.0 cm, locate the nodes and the antinodes. How many loops are formed in the vibration? (a) The amplitude of the vibration of the particle at position x is A = (5.00 mm) sin [(1.57 cm–1) x ]. For x = 5.66 cm,

    A = (5.00mm) sin  2  5.66  = (5.00mm) sin  2.5  3      = (5.00mm ) cos

 = 2.50 mm 3

(b) From the given equation, the wave number k = 1.57 cm–1 and the angular frequency  = 314 s–1. Thus, the wavelength is =

2  3.14 2 = = 4.00 cm 1.57 cm 1 k

 314s 1  = 50 s–1. 2 1.57cm 1 The wave speed is v = v = (50 s–1) (4.00 cm) = 2.00 m/s. (c) The velocity of the particle at position x at time t is given by and the frequency is v =

y = (5.00 mm) sin [(1.57 cm–1) x ] t [314 s–1 cos (314 s–1)t] = (157 cm/s) sin (1.57 cm–1) x cos (314 s–1) t. Putting x = 5.66 cm and t = 2.00 s, the velocity of this particle at the given instant is

v=

 5     cos (200 ) (157 cm/s) sin   2 3  × 1 = 78.5 cm/s. 3 (d) the nodes occur where the amplitude is zero i.e., sin (1.57 cm–1) x = 0.

= (157 cm/s) × cos

or,

   cm 1  x = n, 2 

where n is an integer. Thus, x = 2 n cm. The nodes, therefore, occur at x = 0,2 cm, 4 cm, 6 cm, 8 cm and 10 cm. Antinodes occur in between them i.e., at x = 1 cm, 3 cm, 5 cm, 7 cm and 9 cm. The string vibrates in 5 loops. 8. Sol.

A guitar string is 90 cm long and has a fundamental frequency of 124 Hz. Where should it be pressed to produce a fundamental frequency of 186 Hz? The fundamental frequency of a string fixed at both ends is given by manishkumarphysics.in

Page # 11

Chapter # 15


1 v = 2L

F . 

v1 L 2 As F and  are fixed, v  L 2 1 v1 124 Hz L2 = v L1 = 186 Hz (90 cm) = 60 cm. 2 Thus, the string should be pressed at 60 cm from an end. or,

9.

Sol.

A sonometer wire has a total length of 1m between the fixed ends. Where should the two bridges be placed below the wire so that the three segments of the wire have their fundamental frequencies in the ration 1 : 2 : 3? Suppose the lengths of the three segments are L1, L2 and L3 respectively. The fundamental frequencies are v1 =

1 F 2 L1 

v3 =

1 2L3



v2 =

1 2L2

F 

F 

so that v1 L1 = v2 L2 = v3 L3 . ....(i) As v1 : v2 : v3 = 1 : 2 : 3, we have v2 = 2 v1 and v3 = 3 v1 so that by (i)

v1 L1 L2 = v L1 = 2 2 v1 L1 L3 = v L1  3 3 As L1 + L2 + L3 = 1 m, and

1 1  we get L1 1    = 1m 2 3 

or,

L1 =

6 m. 11

Thus,

L2 =

L1 3 = m 2 11

one bridge should be placed at

and

L3 =

L1 2  m. 3 11

6 2 m from one end and the other should be placed at m from the other 11 11

end. 10.

Sol.

A wire having a linear mass density 5.0 × 10–3 kg/m is stretched between two rigid supports with a tension of 450 N. The wire resonates at a frequency of 420 Hz. The next higher frequency at which the same wire resonates is 490 Hz. Find the length of the wire. Suppose the wire vibrates at 420 Hz in its nth harmonic and at 490 Hz in its (n + 1)th harmonic.

n 420 s–1 = 2 L and

490 s–1 =

(n  1) 2L

F 

....(i)

F 

....(ii)

490 n 1 = 420 n or, n = 6. Putting the value in (i),

This gives


Page # 12

Chapter # 15

Wave Motion and Waves on A string 450 N

6 420 s–1 = 2 L or,

5.0  10 3 kg / m

=

900 m/s L

900 m = 2.1 m 420

L=

EXERCISE 1.

2.

A wave pulse passing on a string with a speed of 40 cm/s in the negative x-direction has its maximum at x = 0 at t = 0. Where will this maximum be located at t = 5 s ? ,d Mksjh eas ,d rjax Lian _.kkRed x- fn'kk esa 40 cm/s pky ls xeu dj jgk gS] t = 0 ij bldk mfPp"B x = 0 ij gSA t = 5 lsd.M ij ;g mfPp"B dgk¡ ij gksxh\ 1. At x = – 2m The equation of a wave travelling on a string stretched along the X-axis is given by t  x    a T   Ae

2

y= (a) Write the dimensions of A, a and T. (b) Find the wave speed. (c) In which direction is the wave travelling? (d) Where is the maximum of the pulse located at t = T? At t = 2T? 2. (a) L, L, T (b) a/T (c) negative x-direction (d) x = – a and x = – 2a X-v{k

ds vuqfn'k ruh gqbZ Mksjh esa xeu dj jgh rjax dh lehdj.k y=

t  x    a T   Ae

2

(a) A, a rFkk T dh foek,¡ fyf[k;sA (b) rjax dh pky Kkr dhft;sA (c) rjax fdl fn'kk esa xeu dj jgh gS ? (d) t = T? At t = 2Tij vf/kdre Lian fdl

LFkkuksa ij gksxsaA

3.

Figure shows a wave pulse at t = 0. The pulse moves to the right with a speed of 10 cm/s. Sketch the shape of the string at t = 1 s, 2 s and 3 s. fp=k esa t = 0 ij ,d rjax Lian n'kkZ;k x;k gSA Lian nka;h vksj 10 cm/s pky ls xfr'khy gSA t = 1 lsd.M, 2 lsd.M rFkk 3 lsd.M ij Mksjh dh vkd`fr cukb;sA

4.

A pulse travelling on a string is represented by the function y=

a3 ( x  vt )2  a 2

,

where a = 5 mm and v = 20 cm/s. Sketch the shape of the string at t = 0, 1 s and 2s. Take x = 0 middle of the string.

in the

fdlh Mksjh esa xeu dj jgk Lian fuEu Qyu }kjk O;Dr fd;k tkrk gSA % y=

a3 ( x  vt ) 2  a 2

,

tgk¡ a = 5 feeh rFkk v = 20 cm/s gSA t = 0, 1 s rFkk 2s ij Mksjh dh vkd`fr dk js[kkfp=k cukb;sA Mksjh ds e/; fcUnq ij x = 0 eku fyft;sA 5.

The displacement of the particle at x = 0 of a stretched string carrying a wave in the positive x-direction is given by f(t) = A sin (t/T). The wave speed is v. Write the wave equation. manishkumarphysics.in

Page # 13

Chapter # 15


,d ruh gqbZ Mksjh esa /kukRed x-fn'kk esa xeu dj jgh rjax ds dkj.k x= 0 ij d.k dk foLFkkiu f(t) = A sin (t/T) }kjk O;Dr fd;k tkrk gSA rjax dh pky v gSA rjax lehdj.k fyf[k;sA [HCV_Ch-15_Ex_5] 5. 6.

7.

8.

x  t  f(x, t) = A sin    T vT 

A wave pulse is travelling on a string with a speed v towards the positive X-axis. The shape of the string at t = 0 is given by g(x) = A sin (x/a), where A and a are constants. (a) What are the dimensions of A and a? (b) Write the equation of the wave for a general time t, if the wave speed is v. [HCV_Ch-15_ExI_6] ,d Mksjh esa rjax Lian /kukRed X-v{k ds vuqfn'k v pky ls xeu dj jgk gSA t = 0 ij Mksjh dh vkd`fr g(x) = A sin (x/ a), }kjk O;Dr dh tkrh gS] tgk¡ A rFkk a fu;rkad gSA (a) A rFkk afu;rkad ? (b) ;fn rjax dh pky v gS rks fdlh le; t ij rjax dh lehdj.k fyf[k;sA x  vt a A wave propagates on a string in the positive x-direction at a velocity v. The shape of the string at t = t0 is given by g(x, t0) = A sin(x/a). Write the wave equation for a general time t. fdlh Mksjh esa ,d rjax v pky ls /kukRed x-fn'kk esa xeu dj jgh gSA t = t0 ij Mksjh dh vkd`fr g(x, t0) = A sin(x/a) }kjk O;Dr dh tkrh gSA fdlh le; t ds fy;s rjax lehdj.k fyf[k;sA

6.

(a) L, L

(b) f(x, t) = A sin

7.

f(x, t) = A sin

x  v (t  t 0 ) a

The equation of a wave travelling on a string is y = (0.10 mm) sin (31.4 m–1)x + 314 s–1)t ] (a) In which direction does the wave travel? (b) Find the wave speed, the wavelength and the frequency of the wave. (c) What is the maximum displacement and the maximum speed of a portion of the string ? 8. (a) negative x-direction (b) 10 m/s, 20 cm, 50 Hz (c) 0.10 mm, 3.14 cm/s

Mksjh esa xeu dj jgh rjax dh lehdj.k fuEu gS % y = (0.10 mm) sin (31.4 m–1)x + 314 s–1)t ] fdl fn'kk esa xeu djsxh ?

(a) rjax (b) rjax ds fy;s] rjax dh pky] rjaxnS/;Z rFkk vko`fr Kkr dhft;sA (c) Mksjh ds fdlh Hkkx dk] vf/kdre foLFkkiu rFkk vf/kdre pky fdruh gks ldrh gS ? 8. (a) negative x-direction (b) 10 m/s, 20 cm, 50 Hz (c) 0.10 mm, 3.14 cm/s 9.

A wave travels along the positive x-direction with a speed of 20 m/s. The amplitude of the wave is 0.20 cm and the wavelength 2.0 cm. (a) write a suitable wave equation which describes this wave. (b) What is the displacement and velocity of the particle at x = 2.0 cm at time t = 0 according to the wave equation written? Can you get different values of this quantity if the wave equation is written in a different fashion? ,d rjax /kukRed x-fn'kk esa 20 m/s pky ls xeu dj jgh gSA rjax dk vk;ke 0.20 cm rFkk rjaxnS/;Z 2.0 cm. (a) bl rjax dks O;Dr djus okyh mfpr rjax lehdj.k fyf[kusA (b) fyf[k gqbZ lehdj.k ds vuqlkj le; t = 0 ij] x = 2.0 lseh ij fLFkr d.k dk foLFkkiu ,oa osx fdruk gksxk \

;fn rjax lehdj.k fdlh vU; :i esa fy[kh tkrh rks D;k vki bu jkf'k;ksa ds fHkUu eku izkIr djrs\a 9. 10.

(a)

y = (0.20 cm) sin [( cm–1) x – (2 × 103 s–1)]

(b) zeor, 4 m/s

A wave is described by the equation

 x t   .  y = (1.0 mm) sin    2.0 cm 0.01s  (a) Find the time period and the wavelength. (b) write the equation for the velocity of the particles. Find the speed of the particles at x = 1.0 cm at time t = 0.01 s. (c) what are the speeds of the particles at x = 3.0 cm, 5.0 cm and 7.0 cm at t = 0.01 s ? (d) What are the speeds of the particles at x = 1.0 cm at t = 0.011, 0.012, and 0.013 s? 10. (a) 20 ms, 4.0 cm (b) zero (c) zero (d) 9.7 cm/s, 18 cm/s, 25 cm/s

,d rjax fuEu lehdj.k }kjk O;Dr dh tkrh gS % manishkumarphysics.in

Page # 14

Chapter # 15


 x t   .  y = (1.0 mm) sin    2.0 cm 0.01s  (a) vkorZdky

rFkk rjaxnS/;Z Kkr dhft;sA (b) d.k ds osx dh lehdj.k fyf[k;sA le; t = 0.01 lsd.M ij x = 1.0 lseh ij fLFkr d.k dh pky Kkr dhft;sA (c) le; t = 0.01 lsd.M ij x = 3.0 lseh, 5.0 lseh rFkk 7.0 lseh ij fLFkr d.k dh pky Kkr dhft;sA (d) x = 1.0 lseh ij fLFkr d.kksa dh pkysa t = 0.011, 0.012, rFkk 0.013 lsd.M ij D;k gksxh \ 10. 11.

(a) 20 ms, 4.0 cm

(b) zero

(c) zero

(d) 9.7 cm/s, 18 cm/s, 25 cm/s

A particle on a stretched string supporting a travelling wave, takes 5.0 ms to move from its mean position to the extreme position. The distance between two consecutive particles, which are at their mean positions, is 2.0 cm. Find the frequency, the wavelength and the wave speed. ,d ruh gqbZ Mksjh dk d.k] ftlesa ,d rjax xeu dj jgh gS] ek/; fLFkfr ls vafre fLFkfr rd xfr djus esa 5.0 feyh lsd.M yxkrk gSA ek/; fLFkfr;ksa ij fLFkr nks Øekxr d.kksa ds e/; dh nwjh 2.0 lseh gSA vko`fr rjaxnS/;Z rFkk rjax dh pky Kkr

dhft;sA 11. 12.

Figure shows a plot of the transverse displacements of the particles of a string at t = 0 through which a travelling wave is passing in the positive x-direction. The wave speed is 20 cm/s. Find (a) the amplitude,(b) the wavelength, (c) the wave number and (d) the frequency of the wave. ,d Mksjh esa ,d rjax /kukRedx-fn'kk dh vksj xeu dj jgh gS] fp=k esa t = 0 ij blds d.kks ds vuqiLz Fk foLFkkiuksa n'kkZrk x;kk gSA rjax dh pky 20 cm/s gSA Kkr dhft;s % (a) vk;ke,(b) rjaxnS/;Z, (c)rjax la[;k rFkk (d) rjax dh vko`fÙk

12. 13.

50 Hz, 4.0 cm, 2.0 m/s

(a)

1.0 mm

(b)

4 cm

(c)

1.6 cm–1

(d)

5 Hz

A wave travelling on a string at a speed of 10 m/s causes each particle of the string to oscillate with a time period of 20 ms. (a) What is the wavelength of the wave ? (b) If the displacement of a particle is 1.5 mm at a certain instant, what will be the displacement of a particle 10 cm away from it at the same instant ? ,d Mksjh esa 10 m/s pky ls xeu dj jgh rjax ds dkj.k Mksjh dk izR;sd d.k 20 feyh lsd.M vkorZdky ds lkFk nksyu djrk gSA (a) rjax dh rjaxnS/;Z fdruh gS\ (b) ;fn fdlh {k.k ij ,d dk foLFkkiu 1.5 feeh gS rks blh {k.k ij blls 10

lseh nwj fLFkr fdlh d.k dk foLFkkiu fdruk gksxk \ 14.

13. (a) 20 cm (b) – 1.5 mm A steel wire of length 64 cm weighs 5 g. If it is stretched by a force of 8 N, what would be the speed of a transverse wave passing on it? 64 lseh yEcs d.k ij LVhy ds rkj dk Hkkj 5 xzke gSA ;fn bldks 8 N cy ls rkuk tk;s] blls xqtjus okyh vuqiLz Fk rjax

dh pky fdruh gksxh \ 14. 15.

A string of length 20 cm and linear mass density 0.40 g/cm is fixed at both ends and is kept under a tension of 16 N. A wave pulse in produce at t = 0 near an end as shown in figure, which travels towards the other end. (a) When will the string have the shape shown in the figure again? (b) Sketch the shape of the string at a time half of that found in part (a). [HCV_Ch-15_Ex_15] 20 lseh yEch rFkk 0.40 xzke@lseh jSf[k; nzO;eku ?kuRo okyh Mksjh dks nksuksa fljks ij dldj 16 U;wVu ruko ls rkuk x;k gSA fp=kkuqlkj t = 0 ij blds ,d fljs ij ,d rjax Lian mRiUu fd;k x;k gS] tks nwljs fljs dh vksj xeu dj jgk gSA (a) Mksjh dh vkd`fr iqu% fp=kkuqlkj dc gksxh \ (b) Hkkx (a) esa izkIr le; ls vk/ks le; ij Mksjh dh vkd`fr dk js[kkfp=k cukb;sA

15. 16.

32 m/s

(a) 0.02 s

A string of linear mass density 0.5 g/cm and a total length 30 cm is tied to a fixed wall at one end and to a frictionless ring at the other end (figure). The string can move on a vertical rod. A wave pulse is produced on manishkumarphysics.in

Page # 15

Chapter # 15 Wave Motion and Waves on A string the string which moves towards the ring at a speed of 20 cm/s. The pulse is symmetric about its maximum which is located at a distance of 20 cm from the end joined to the ring. (a) Assuming that the wave is reflected from the ends without loss of energy, find the time taken by the string to regain its shape. (b) The shape of the string changes periodically with time. Find this time period (c) What is the tension in the string ? ,d Mksjh dk jSf[k; nzO;eku ?kuRo 0.5 xzke@lseh gS rFkk bldh dqy yEckbZ 30 lseh gS] bldk ,d fljk nhokj esa dlk gqvk

gS rFkk nwljk fljk ,d ?k"kZ.k jfgr oy; ls ca/kk gqvk gSA ¼fp=k½] oy; m/okZ/kj NM+ ij xfr dj ldrh gSA Mksjh esa ,d rjax&Lian mRiUu fd;k x;k gS tks oy; dh vksj 20 lseh@ls pky ls xeu dj jgk gS \ Lian oy; ls ca/ks gq, fljs ls 20 lseh nwj gS rFkk blds mPpre fcUnq ds nksuks vksj lefer gsA (a) ;g eku yhft;s fd rajx fljksa ij fcuk ÅtkZ gkfu ds ijkofrZr gksrh gS] Mksjh dks iqu% ;gh vkd`fr izkIr djus esa yxus okyk l;e Kkr dhft;sA (b) Mksjh dh vko`fr le; ds lkFk vkorhZ :i ls ifjofrZr gksrh gS] vkorZdky Kkr dhft;sA (c) Mksjh esa ruko fdruk gS \ [HCV_Ch-15_Ex_16]

16. 17.

(a)

2s

(b)

3s

(c)

2 × 10–3 N

Two wires of different densities but same area of cross-section are soldered together at one end and are stretched to a tension T. The velocity of a transverse wave in the first wire is double of that in the second wire. Find the ratio of the density of the first wire to that of the second wire. fHkUu&fHkUu ?kuRo fdUrq leku vuqiLz Fk dkV {ks=kQy okys nks rkj ,d fljs ij >kya dj tksMs x;s gS (soldered) rFkk T ruko

ls rkus x;s gSA izFke rkj esa ,d vuqiLz Fk rjax dk osx] nwljs rkj esa osx dk nqxuk gSA izFke rkj ds ?kuRo ls nljs rkj ds ?kuRo dk vuqikr Kkr dhft;sA 17. 18.

0.25

A transverse wave described by y = (0.02 m) sin [(1.0 m–1) x + (30 s–1) t] propagates on a stretched string having a linear mass density of 1.2 × 10–4 kg/m. Find the tension in the string.

,d vuqizLFk rjax fuEukuqlkj O;Dr dh tkrh gS% y = (0.02 eh-) sin [(1.0 eh-–1) x + (30 ls-–1) t] ;g ,d ruh gqbZ Mksjh esa xeu djrh gSA ftldk jSf[kd nzO;eku ?kuRo 1.2 × 10–4 fdxzkk@eh- gSA Mksjh esa ruko Kkr dhft;sA 18. 19.

0.108 N

A travelling wave is produced on a long horizontal string by vibrating an end up and down sinusoidally. The amplitude of vibration is 1.0 cm and the displacement becomes zero 200 times per second. The linear mass density of the string is 0.10 kg/m and it is kept under a tension of 90N. (a) Find the speed and the wavelength of the wave. (b) Assume that the wave moves in the positive x-direction and at t = 0, the end x = 0 is at its positive extreme position. Write the wave equation. (c) Find the velocity and acceleration of the particle at x = 50 cm at time t = 10 ms.

,d yEch {ksfrt Mksjh ds ,d fljks dks T;koØh; :i ls Åij uhps dEiUu djokdj blesa ,d izxkeh rjax mRiUu dh tkrh gSA dEiUu dk vk;ke 1.0 lseh gS rFkk foLFkkiu izfr lsd.M 200 ckj 'kwU; gks tkrk gSA Mksjh dk jSf[kd nzO;eku /kuRo 0.10 fdxzk@eh gS rFkk blesa ruko 90 U;wVu gSA (a) rjax dh pky rFkk rjax ruko nS/;Z Kkr dhft;sA (b)eku yhft;s fd rjax /kukRed x &fn'kk esa xeu dj jgh gS rFkk t = 0 ij bldk x = 0 fljk /kukRed vf/kdre fLFkfr esa gSA rjax lehdj.k fyf[k;sA (c) le; t == 10 feyh lsd.M+ ij x = 50 lseh ij fLFkr d.k dk osx rFkk Roj.k Kkr dhft;sA 19.

20.

(a)

30 m/s, 30 cm (b)

(c)

– 5.4 m/s, 2.0 km/s2

 x t  y = (1.0 cm) cos 2  30 cm  0.01s   

A string of length 40 cm and weighing 10 g is attached to a spring at one end and to a fixed wall at the other end. The spring has a spring constant of 160 N/m and is stretched by 1.0 cm. If a wave pulse is produced on the string near the wall, how much time will it take to reach the spring? HCV_Ch-15_ExI_20 40 lseh yEch rFkk 10 xzke Hkkj okyh Mksjh dk ,d fljk fLizx a ls rFkk nwljk fljk nhokj ls tqM+k gqvk gSA fLizxa fu;rkad 160 U;wVu@eh- gS rFkk ;g 1.0 lseh izlkfjr gSA ;fn Mksjh esa nhokj ds lehi rjaxLian mRiUu fd;k tk;s rks ;g fLizxa rd fdrus

le; esa igqpsxk\ Ans.

0.05 s


Page # 16

Chapter # 15 Wave Motion and Waves on A string 21. Two blocks each having a mass of 3.2 kg are connected by a wire CD and the system is suspended from the ceiling by another wire AB (figure). The linear mass density of the wire AB is 10 g/m and that of CD is 8 g/m. Find the speed of a transverse wave pulse produced in AB and in CD ,d rkj CD ls nks xqVds tqMs gq, gS] izR;sd dk nzO;eku3.2 fdxzk gS rFkk ;g fudk; ,d vU; rkj AB }kjk Nr ls yVdk;k x;k gS (fp=k) rkj AB dk jSf[kd ?kuRo 10 xzke@eh- rFkk CD dk 8 xzke@eh- gSA AB ,oa CD esa mRiUu vuqizLFk rjax Lian dh

pky Kkr dhft;sA

21.

22.

In the arrangement shown in figure, the string has mass of 4.5 g. How much time will it take for a transverse disturbance produced at the floor to reach the pulley ? Take g = 10 m/s2. fp=k esa n'kkZ;h x;h O;oLFkk es]a Mksjh dk nzO;eku 4.5 g gSA Mksjh esa Q'kZ ds lehi mRiUu vuqçLFk fo{kksHk dks f?kjuh rd igqp ¡ us esa fdruk le; yxsxk ? g = 10 m/s2 ekfu,A [M.Bank(07-08)_HCV_Ch.15_Ex._22]

Ans. Sol.

0.02 s

v=

T 

t=

2 = v

t=2

23.

79 m/s and 63 m/s

2 T

 =2

m LMg

4.5 10 3 = 0.02 sec. (2.25) (2) (10)

A 4.0 kg block is suspended from the ceiling of an elevator through a string having a linear mass density of 19.2 × 10–3 kg/m. Find the speed (with respect to the string) with which a wave pulse can proceed on the string if the elevator accelerates up at the rate of 2.0 m/s2. Take g = 10 m/s2. ,d fy¶V dh Nr ls 19.2 × 10–3 fdxzk@eh- jSf[kd ?kuRo okyh Mksjh dh lgk;rk ls ,d 40 fdxzk nzO;eku dk xqVdk yVdk;k x;k gSA ;fn fy¶V 2.0 eh-@ls-2 dh nj Åij dh vksj Rofjr gks jgh gks rks Mksjh esa ,d rjax Lian ds xeu dh pky¼Mksjh

ds lkis{k½ Kkr dhft;sA 23. 24.

50 m/s

A heavy ball is suspended from the ceiling of a motor car through a light string. A transverse pulse travels at a speed of 60 cm/s on the string when the car is at rest and 62 cm/s when the car accelerates on a horizontal road. Find the acceleration of the car. Take g = 10 m/s2.

,d eksVj dkj dh Nr ls ,d gYdh Mksjh dh lgk;rk ls ,d Hkkjh xsna yVdk;h x;h gSA tc dkj fojkekoLFkk esa gS rks ,d vuqiLz Fk Lian Mksjh esa 60 lseh@ls pky ls rFkk tc dkj {ksfrt lM+d ij Rofjr gS rks 62 lseh@ls pky ls xeu djrk gSA dkj dk Roj.k Kkr dhft;sA ¼ g = 10 eh-@ls-2) 24. 25.

3.7 m/s2

A circular loop of string rotates about its axis on a frictionless horizontal plane at a uniform rate so that the tangential speed of any particle of the string is v. If a small transverse disturbance is produced at a point of the loop, with what speed (relative to the string) will this disturbance travel on the string ? manishkumarphysics.in

Page # 17

Chapter # 15


,d {ksfrt ,oa ?k"kZ.k jfgr lery ij ,d Mksjh dk o`Ùkkdkj ywi viuh v{k ds ikfjr% ,d leku nj ls ?kw.kZu dj jgk gS]ftlls Mksjh ds fdlh Hkh d.k dh Li'kZ js[kh; pky v gSA ;fn ywi ds fdlh fcUnq ij ,d vYi vuqiLz Fk fo{kksHk mRiUu fd;k tk;s] rks ;g fo{kksHk Mksjh ij fdl pky ls ¼Mksjh ds lkis{k ½ xeu djsxk \ 25.

v

26.

A heavy but uniform rope of length L is suspended from a ceiling. (a) Write the velocity of a transverse wave travelling on the string as a function of the distance from the lower end. (b) If the rope is given a sudden sideways jerk at the bottom, how long will it take for the pulse to reach the ceiling? (c) A particle is dropped from the ceiling at the instant the bottom end is given the jerk. Where will the particle meet the pulse? L 26. (a) gx (b) (c) at a distance from the bottom 4L / g 3 L yEckbZ dh ,d Hkkjh fdUrq ,d leku jLlh Nr ls yVdk;h x;h gSA (a) Mksjh esa xeu djus okys vuqiLz Fk ladr s dh pky] Mksjh ds fupys fljs ls nwjh ds Qyu ds :i esa fyf[k;sA (b) ;fn Mksjh dks uhps vpkud cxy esa >Vdk fn;k tk;s] Nr rd ladsr iagqpus esa fdruk le; yxsxk \ (c) A particle is dropped from the ceiling at the instant the bottom end is given the jerk. Where will the particle meet the pulse? L 26. (a) gx (b) (c) at a distance from the bottom 4L / g 3

27.

Two long strings A and B, each having linear mass density 1.2 × 10–2 kg/m, are stretched by different tensions 4.8 N and 7.5 N respectively and are kept parallel to each other with their left ends at x = 0. Wave pulses are produced on the strings at the left ends at t = 0 on string A and at t = 20 ms on string B. When and where will the pulse on B overtake that on A? nks yEch Mksfj;kW A rFkk B izR;sd dk jSf[kd nzO;eku ?kuRo 1.2 × 10–2 fdxzk@eh- gS] fHkUu&fHkUu rukoksa Øe'k% 4.8 U;wVu ,oa 7.5 U;wVu ls ruh gqbZ gS rFkk Mksfj;kW budk cka;k fljk x = 0 ij j[krs gq, ijLij lekukarj j[kh x;h gSA Mksjh A esa t = 0 ij rFkk B esa t = 20 feyh lsd.M ij Mksfj;kW ds ck;sa fljs ij vuqizLFk Lian mRiUu fd;s tkrs gSA dc rFkk fdl LFkku ij Mksjh B dk Lin Mksjh A ds Lian ls vkxs fudy tk;sxk \ 27.

at t = 100 ms at x = 2.0 m

28.

A transverse wave of amplitude 0.50 mm and frequency 100 Hz is produced on a wire stretched to a tension of 100 N. If the wave speed is 100 m/s, what average power is the source transmitting to the wire? 100 U;wVu ruko ls rus gq, rkj esa 0.50 feeh vk;ke rFkk 100 gV~Zt vko`fr dh vuqizLFk rjax mRiUu dh tkrh gSA ;fn rjax dh pky 10 eh@ls gS] L=kksr }kjk rkj dks vkSlr fdruh 'kfDr LFkkukarfjr dh tk jgh gSA 28. 49 mW

29.

A 200 Hz wave with amplitude 1 mm travels on a long string of linear mass density 6 g/m kept under a tension of 60 N. (a) Find the average power transmitted across a given point on the string. (b) Find the total energy associated with the wave in a 2.0 m long portion of the string. [HCV_Ch-15_Ex_29] 6 g/m nzO;eku ?kuRo rFkk 60 N ruko ls ruh gqbZ ,d yEch Mksjh ij 200 Hz vko`fr rFkk 1 mm vk;ke okyh rjax xfreku gSA (a) Mksjh ds fdlh fn;s x;s fcUnq ls lapfjr vkSlr 'kfDr Kkr dhft,A (b) Mksjh ds 2.0 m Hkh yEcs Hkkx esa rjax ls lEc)

dqy ÅtkZ Kkr dhft,A 29. 30.

(a)

0.47 W

(b)

9.4 mJ

A tuning fork of frequency 440 Hz is attached to a long string of linear mass density 0.01 kg/m kept under a tension of 49 N. The fork produces transverse waves of amplitude 0.50 mm on the string. (a) Find the wave speed and the wavelength of the waves. (b) Find the maximum speed and acceleration of a particle of the string. (c) At what average rate is the tuning fork transmitting energy to the string? 49 U;wVu ruko ls ruh gqbZ rFkk 0.01 fdxzk@eh jSf[kd nzO;eku ?kuRo okyh Mksjh ls 440 gV~t Z vko`fr dk ,d Lofj=k&f}Hkqt tqM+k gqvk gSA Lofj=k Mksjh esa 0.50feeh vk;ke dh vuqiLz Fk rjaxs mRiUu djrk gSA (a) rjaxks dh rjax pky rFkk rjax nS/;Z Kkr dhft;sA (b) Mksjh ds fdlh d.k dh vf/kdre pky ,oa Roj.k Kkr dhft;sA (c) Lofj=k fdl vkSlr nj ls Mksjh dks ÅtkZ

LFkkukarfjr dj jgk gS \ 30.

31.

(a)

70 m/s, 16 cm (b)

1.4 m/s, 3.8 km/s

(c)

0.67 W

Two waves, travelling in the same direction through the same region, have equal frequencies, wavelengths and amplitudes. If the amplitude of each wave is 4 mm and the phase difference between the waves is 90°, what is the resultant amplitude?

,d gh {ks=k esa nks rjaxs ,d gh fn'kk esa xeu dj jgh gS] budh vko`fr;k¡] rjaxnS/;Z ,oa vk;ke ,d leku gSA ;fn izR;sd rjax manishkumarphysics.in

Page # 18

Chapter # 15


dk vk;ke 4 feeh gS rFkk rjaxks ds e/; dykUrj 90° gS] ifj.kkeh vk;ke fdruk gksxk\ 31.

4 2 mm

32.

Figure shows two wave pulses at t = 0 travelling on a string in opposite directions with the same wave speed 50 cm/s. Sketch the shape of the string at t = 4 ms, 8 ms and 12 ms. [HCV_Ch-15_Ex_32] fp=k esa n'kkZ;k x;k gS fd ,d Mksjh esa t = 0 ij nks rjax Lian foifjr fn'kk esa leku rjax pky 50 lseh@ls ls xeu dj jgs gSA t = 4 feeh lssd.M 8 feyh lsd.M ,oa 12 feyh lsd.M ij Mksjh dh vkd`fr dk js[kkfp=k cukb;sA

33.

Two waves, each having a frequency of 100 Hz and a wavelength of 2.0 cm, are travelling in the same direction on a string. What is the phase difference between the waves (a) if the second wave was produced 0.015 s later than the first one at the same place, (b) if the two waves were produced a distance 4.0 cm behind the second one? (c) If each of the waves has an amplitude of 2.0 mm, what would be the amplitudes of the resultant waves in part (a) and (b) ? ,d Mksjh esa ,d gh fn'kk esa nks rajxs xeu dj jgh gS] izR;sd dh vko`fr 100 gV~t Z rFkk rjaxnS/;Z 2.0 lseh gSA rjaxks ds e/ ; dykarj fdruk gksxkA (a) ;fn f}rh; rjaxs 0.015 lsd.M+ i'pkr mRiUu dh xbZ tc izFke mlh LFkku ij FkhA (b) ;fn nksuksa rjaxs ,d gh {k.k ij mRiUu dh x;h fdUrq izFke nwljh ls 4.0 lseh ihNs mRiUu dh x;h\ (c) ;fn fdUrq rjax dk vk;ke 2.0 feeh gS] rks Hkkx (a) rFkk (b) esa ifj.kkeh rjaxksa ds vk;ke fdrus gksxas \ 33. (a) 3 (b) 4 (c) zero, 4.0 mm

34.

If the speed of a transverse wave on a stretched string of length 1 m is 60 m/s, what is the fundamental frequency of vibration? ;fn 1 eh- yEch ruh gqbZ Mksjh esa vuqiLz Fk rjax dh pky 60 eh@ls gS] dEiUu dh ewy vko`fÙk fdruh gksxh \ 34. 30 Hz

35.

A wire of length 2.00 m is stretched to a tension of 160 N. If the fundamental frequency of vibration is 100 Hz, find its linear mass density. 2.00 eh- yEck ,d rkj 160 U;wVu ruko ls [khpk x;k gSA ;fn dEiUu dh ewy vko`fÙk 100 gV~t Z gS] bldk jSf[kd nzO;eku

?kuRo Kkr dhft;sA 35. 36.

1.00 g/m

A steel wire of mass 4.0 g and length 80 cm is fixed at the two ends. The tension in the wire is 50 N. Find the frequency and wavelength of the fourth harmonic of the fundamental. 4.0 xzke nzO;eku rFkk 80 lseh yEckbZ okyk LVhy dk ,d rkj nksuksa fljksa ij dlk gqvk gSA rkj esa 50 U;wVu ruko gSA ewy

ls prqFkZ lUukfn dh vko`fÙk ,oa rjaxnS/;Z Kkr dhft;sA 36. 37.

250 Hz, 40 cm

A piano wire weighing 6.00 g and having a length of 90.0 cm emits a fundamental frequency corresponding to the “Middle C” (v = 261.63 Hz). Find the tension in the wire. fi;kuks ds rkj dk Hkkj 6.00 xzke rFkk yEckbZ 90.0 lseh gS] ;g “e/; C” (v = 261.63 gV~t Z ) ds laxr ,d ewy vko`fÙk mRiUu

dj jgk gSA rkj esa ruko Kkr dhft;sA 37. 38.

1480 N

A sonometer wire having a length of 1.50 m between the bridges vibrates in its second harmonic in resonance with a tuning fork of frequency 256 Hz. What is the speed of the transverse wave on the wire? ,d Lojekih ds rkj dh lsrv q ksa ds e/; yEckbZ 1.50 eh- gS] ;g vius f}rh; lUuknh ls dEiUu djrk gqvk 256 gV~t Z vko`fÙk okys Lofj=k ds lkFk vuqukn mRiUu dj jgk gSA rkj esa vuqiLz Fk rjax dh pky fdruh gS\ [3 min.] [M.Bank(07-08)_HCV_Ch.15__Ex._38] Ans. 384 m/s manishkumarphysics.in

Page # 19

Chapter # 15 Sol.

f=


P T 2 m

1.5m Speed V = 39.

P=2 f = 256 Hz

T 2 2  1.5 =f× = 256 × = 384 m/s. m P 2

The length of the wire shown in figure between the pulleys is 1.5 m and its mass is 12.0 g. Find the frequency of vibration with which the wire vibrates in two loops leaving the middle point of the wire between the pulleys at rest. fp=k esa iznf'kZr rkj dh f?kjfu;ksa ds e/; yEckbZ 1.5 eh rFkk bldk nzO;eku 12.0 xzke gSA nksuksa f/kjfu;ks ds chp dk rkj e?;

fcUnq ij fojkekoLFkk esa jgrs gq, mks ywiksa dEiUu dj jgk gS] dEiUu dh vko`fÙk Kkr dhft;sA

40.

41.

39. 70 Hz A one metre long stretched string having a mass of 40 g is attached to a tuning fork. The fork vibrates at 128 Hz in a direction perpendicular to the string. What should be the tension in the string if it is to vibrate in four loops? 40 xzke nzO;eku dh ,d ehVj yEch ruh gqbZ Mksjh ,d yEdor~ fn'kk esa 128 gV~t Z vko`fÙk ls dEiUu dj jgk gSA ;fn ;g 4 ywik esa dEiUu dj jgh gS] Mksjh esa ruko fdruk gS ? 40. 164 N A wire, fixed at both ends is seen to vibrate at a resonant frequency of 240 Hz and also at 320 Hz. (a) what could be the maximum value of the fundamental frequency ? (b) If transverse waves can travel on this string at a speed of 40 m/s, what is its length ? [HCV_Ch-15_ExI_41] nksuksa fljksa ij dlk gqvk ,d rkj 240 gV~t Z vuqukfn vko`fÙk rFkk lkFk gh 320 gV~t Z vko`fÙk ds lkFk dEiUu djrk gqvk ns[kk tkrk gSA (a) ewy vko`fÙk dk vf/kdre eku fdruk gks ldrk gS ? (b) ;fn bl Mksjh ij vuqiLz Fk rjax 40 eh@ls pky ls xeu djrh gS] bldh yEckbZ fdruh gS ? 41.

(a)

80 Hz

(b)

25 cm

42.

A string, fixed at both ends, vibrates in a resonant mode with a separation of 2.0 cm between the consecutive nodes. For the next higher resonant frequency, this separation is reduced to 1.6 cm. Find the length of the string. [HCV_Ch-15_ExI_42] nksuksa fljksa ij dlh gqbZ Mksjh vuqukfn fo/kk esa bl izdkj dEiUu dj jgh gSA fd nks Øekxr fuLianks ds e?; 2.0 lseh varjky gSA vxyh mPp vuqukfn vko`fÙk ds fy;s] ;g varjky de djds 1.6 lseh dj fn;k x;k gSA Mksjh dh yEckbZ Kkr dhft;sA 42. 8.0 cm

43.

A 660 Hz tuning fork sets up vibration in a string clamped at both ends. The wave speed for a transverse wave on this string is 220 m/s and the string vibrates in three loops. (a) Find the length of the string. (b) If the maximum amplitude of a particle is 0.5 cm, write a suitable equation describing the motion. 660 gV~t Z vko`fÙk okyk ,d Lofj=k f}Hkqt] nksuksa fljksa ij dlh gqbZ Mksjh esa dEiUu mRiUu dj jgk gSA bl Mksjh esa vuqiLz Fk rjax ds fy;s rjax pky 220 eh-@ls gS rFkk Mksjh rhu ywiks esa dEiUu dj jgh gSA (a) Mksjh dh yEckbZ Kkr dhft;s (b) ;fn fdlh d.k dk vf/kdre vk;ke 0.5 djus ds fy;s mfpr lehdj.k fyf[k;sA 43. (a) 50 cm (b) (0.5 cm) sin [(0.06  cm–1)x] × cos[(1320  s–1)t]

44.

A particular guitar wire is 30.0 cm long and vibrates at a frequency of 196 Hz when no finger is placed on it. The next higher notes on the scale are 220 Hz, 247 Hz, 262 Hz and 294 Hz. How far from the end of the string must the finger be placed to play these notes? ,d fof'k"V fxVkj dk rkj 30.0 lseh yEck gS rFkk tc bl ij dksbZ Åaxyh j[kh gqbZ ugh gSA 196 gV~t Z vko`fÙk ls dEiUu dj jgk gSA iSekus 220 gV~t Z , 247 gV~t Z , 262 gV~t Z rFkk 294 gV~t Z gSA bu vko`fr;ksa (notes) dks ctkus ds fy;s fljs ls fdruh

nwjh ij Åaxyh j[kuh gksxh \


Page # 20

Chapter # 15 Wave Motion and Waves on A string 44. 26.7 cm, 23.8 cm, 22.4 cm and 20.0 cm 45.

A steel wire fixed at both ends has a fundamental frequency of 200 Hz. A person can hear sound of maximum frequency 14 kHz. What is the highest harmonic that can be played on this string which is audible to the person? nksuksa fljksa ls dls gq, ,d LVhy ds rkj dh ewy vko`fÙk 200 gV~t Z gSA ,d O;fDr vf/kdre 14 fdyks gV~Zt vko`fÙk dh /

ouh lqu lds\ 45. 46.

70

Three resonant frequencies of a string are 90, 150 and 210 Hz. (a) Find the highest possible fundamental frequency of vibration of this string. (b) Which harmonics of the fundamental are the given frequencies ? (c) Which overtone are these frequencies. (d) If the length of the string is 80 cm, what would be the speed of a transverse wave on this string ? [HCV_Ch-15_Ex_46] ,d Mksjh dh rhu vuqukfn vko`fÙk;k¡ 90, 150 ,oa 210 gV~t Z gSA (a) bl Mksjh ds dEiUu dh vf/kdre lEHko ewy vko`fÙk Kkr dhft;sA (b) nh xbZ vko`fÙk;k¡ ewy dh dkSulh lUukfn gS\ (c) ;g vko`fÙk;k¡ dkSuls lUukn gSA (d) ;fn Mksjh dh yEckbZ 80 lseh

gS] bl Mksjh esa vuqiLz Fk rjax dh pky fdruh gS\ 46.

(a)

30 Hz

(b)

3rd, 5th and 7th (c)

47.

Two wires are kept tight between the same pair of supports. The tensions in the wires are in the ratio 2 : 1, the radii are in the radio 3 : 1 and the densities are in the ratio 1 : 2. Find the ratio of their fundamental frequencies. nks rkj] lgkjksa ds ,d gh tksM+s ds e/; dls gq, gSA rkjksa esa rukokas dk vuqikr 2 : 1, f=kT;kvksa dk vuqikr 3 : 1 rFkk ?kuRoksa dk vuqikr 1 : 2 gSA budh ewy vko`fÙk;k¡ dk vuqikr Kkr dhft;sA 47. 2:3

48.

A uniform horizontal rod of length 40 cm and mass 1.2 kg is supported by two identical wires as shown in figure. Where should a mass of 4.8 kg be placed on the rod so that the same tuning fork may excite the wire on left into its fundamental vibrations and that on right into its first overtone? Take g = 10 m/s2. 40 lseh yEch rFkk 1.2 fdxzk nzO;eku okyh ,d le:i NM+ fp=kkuqlkj nks ,d leku rkjksa ds lgkjs ls yVdkbZ xbZ gSA NM+ ij ,d 4.8 fdxzk nzO;eku dgk¡ ij j[kk tk;s fd ,d gh Lofj=k f}Hkqt ck;sa rkj dks ewy fo/kk esa rFkk nk;sa rkj dks izFke lUukn esa dfEir djsa\ (g = 10 m/s2)

48. 49.

Figure shows an aluminium wire of length 60 cm joined to a steel wire of length 80 cm and stretched between two fixed supports. The tension produced in 40 N. The cross-sectional area of the steel wire is 1.0 mm2 and that of the aluminium wire is 3.0 mm2. What could be the minimum frequency of a tuning fork which can produce standing waves in the system with the joint as a node? The density of aluminium is 2.6 g/cm3 and that of steel is 7.8 g/cm3. fp=k esa n'kkZ;k x;k gS fd 60 lseh yEck rkj LVksy ds 80 lseh yEcs rkj ls tksMk+ x;k gSA rFkk nks fLFkj lgkjs ds e/; rkuk x;k gSA mRiUu fd;k x;k ruko 40 U;wVu gSA LVhy ds rkj dh vuqiLz Fk dkV dk {ks=kQy 1.0 mm2 gS rFkk ,Y;wfefu;e ds rkj dk 3.0 mm2 gSA fdlh Lofj=k dh U;wure vko`fÙk fdruh gksuh pkfg;s fd og bl fudk; esa laf/k ij fuLian j[krs gq, vizxkeh rjaxs mRiUu dj lds\,Y;wfefu;e dk ?kuRo 2.6 g/cm3 rFkk LVhy dk 7.8 g/cm3 gSA

49. 50.

5 cm from the left end

180 Hz

A string of length L fixed at both ends vibrates in its fundamental mode at a frequency v and a maximum amplitude A. (a) Find the wavelength and the ware number k. (b) Take the origin at one end of the string and X-axis along the string. Take the Y-axis along the direction of the displacement. Take t = 0 and the instant when the middle point of the string passes through its mean position and is going towards the positive ydirection. Write the equation describing the standing wave. [HCV_Ch-15_Ex_50] L yEckbZ dh ,d Mksjh nksuksa fljksa ij dlh gqbZ gS rFkk bldh ewy fo/kk esa v vko`fÙk v ,oa vf/kdre vk;ke A ls dEiUu dj jgh gSA (a) rjaxnS/;Z rFkk rjax la[;k k Kkr dhft;sA (b) Mksjh ds ,d fljs ij ewy fcUnq rFkk X-v{k Mksjh ds vuqfn'k eku yhft;sA Y-v{k foLFkkiu ds vuqfn'k eku yhft;sA ftl {k.k ij Mksjh dk e/; fcUnq tc ek/; fLFkfr ls xqtj jgk gS rFkk / manishkumarphysics.in

Page # 21

Chapter # 15


kukRed v fn'kk esa tk jgk gSA t = 0 eku yhft;sA 50.

(a)

2L, /L

(b)

y = A sin (x/L) sin (2vb)

51.

A 2 m long string fixed at both ends is set into vibrations in its first overtone. The wave speed on the string is 200 m/s and the amplitude is 0.5 cm. (a) Find the wavelength and the frequency. (b) Write the equation giving the displacement of different points as a function of time. Choose the X-axis along the string with the origin at one end at t = 0 at the instant when the point x = 50 cm has reached its maximum displacement. nksuksa fljksa ij dlh gqbZ 2 eh- yEch ,d Mksjh blds izFke lUukn ls dfEir djokbZ tk jgh gSA Mksjh esa rjax dh pky 200 eh@ls rFkk vk;ke 0.5 leh gSA (a) rjaxnS/;Z rFkk vko`fÙk Kkr dhft;sA (b) fofHkUu fcUnqvksa ij foLFkkiu dh le; ds Qyu ds :i esa lehdj.k fyf[k;sA Mksjh ds vuqfn'k X-v{k rFkk ,d fljs ij ewy fcUnq eku yhft;s ,oa tc x = 50 lseh ij fLFkr fcUnq vf/kdre foLFkkiu dh fLFkfr esa gksrk gSA t = 0 eku yhft;sA 51. (a) 2m, 100 Hz (b) (0.5 cm) sin [(m–1) x] cos [(200 s–1)t]

52.

The equation for the vibration of a string, fixed at both ends vibrating in its third harmonic, is given by y = (0.4 cm) sin [(0.314 cm–1) x] cos [(600  s–1) t]. (a) What is the frequency of vibration ? (b) What are the positions of the nodes ? (c) What is the length of the string ? (d) What is the wavelength and the speed of two travelling waves that can interfere to give this vibration ?

r`rh; lUkkfn esa dEiUu dj jgh] nksuksa fljks ij dlh gqbZ Mksjh ds dEiUu dh lehdj.k fuEukuqlkj O;Dr dh tkrh gSA y = (0.4 cm) sin [(0.314 cm–1) x] cos [(600  s–1) t]. (a) dEiUu dh vko`fÙk D;k gS (b) fuLianksa dh fLFkfr;k¡ D;k gS? (c) Mksjh dh gksdj mRiUu djus okyh nks izxkeh rjaxks dh vko`fÙk ,oa pky D;k gS? 52. (a) 300 Hz (b) 0, 10 cm, 20 cm, 30 cm (c) 53.

yEckbZ fdruh gS? (d) bl dEiUu dks v/;kjksfir 30 cm (D)20 cm, 60 m/s

The equation of a standing wave, produced on a string fixed at both ends, is y = (0.4 cm) sin [(0.314 cm–1) x] cos [(600 s–1) t]. What could be the smallest length of the string ? [HCV_Ch-15_Ex_53]

nksuksa fljks ij dlh gqbZ Mksjh esa vizxkeh rjax dh lehdj.k gS % Mksjh dh 53.

y = (0.4 cm) sin [(0.314 cm–1) x] cos [(600 s–1) t]. U;wure yEckbZ fdruh gS? 10 cm

54.

A 40 cm wire having a mass of 3.2 g is stretched between two fixed supports 40.05 cm apart. In its fundamental mode, the wires vibrates at 220 Hz. If the area of cross-section of the wire is 1.0 mm2, find its Young’s modulus. 32 xzke nzO;eku dk 40 lseh yEck rkj 40.05 lseh nwj nks fLFkj f'kadtks ds e/; [khpa dj dlk x;k gSA bldh ewy fo/kk esa dEiUu vko`fÙk 220 gV~t Z gSA ;fn rkj ds vuqiLz Fk dkV dk {ks=kQy 1.0 mm2 gS] ;ax dk izR;kLFkrk xq.kkad Kkr dhft;sA 54. 1.98 × 1011 N/m2

55.

Figure shows a string stretched by a block going over a pulley. The string vibrates in is length harmonic in unison with a particular tuning fork. When a beaker containing water is brought under the block so that the block is completely dipped into the beaker, the string vibrates in is eleventh harmonic. Find the density of the material of the block.

fp=k esa iznf'kZr Mksjh] f?kjuh ls gksdj yVdk, x;s xqVds }kjk [khaph tk jgh gSA Mksjh blds nlosa lUukfn }kjk [khph tk jgh gSA Mksjh blds nlosa lUukfn esa dEiUu djrh gqbZ ,d fof'k"V Lofj=k f}Hkqt ds lkFk leLOfjr gSA tc xqVds ds uhps ikuh ls Hkjk gqvk ,d ik=k yk;k tkrk gS\ ftlls xqVdk ikuh esa iwjk Mwc tkrk gS] rks Mksjh blds X;kjgosa lUukfn esa dEiUu djrh gSA xqVds ds inkFkZ dk ?kuRo Kkr dhft;sA

55. 56.

5.8 × 113 kg/m3

A 2.00 m long rope, having a mass of 80 g, is fixed at one end and is tied to a light string at the other end. The tension in the string is 256 N. (a) Find the frequencies of the fundamental and the first two overtones. (b) Find the wavelength in the fundamental and the first two overtones. manishkumarphysics.in

Page # 22

Chapter # 15


,d 2.00 eh- yEch Mksjh dk nzO;eku 80 xzke gS] ;g ,d fljs ij dlh gqbZ gS rFkk nwljs fljs ij ,d gYdh Mksjh ls ca/kh gqbZ gSA Mksjh esa 256 U;wVu dk ruko gSA (a) ewy fo/kk rFkk izFke nks lUuknks dh vko`fÙk;k¡ Kkr dhft;sA (b) ewy fo/kk rFkk izFke nks lUuknks dh rjax nS/;Z Kkr dhft;sA 56. 57.

(a)

10 hz, 30 Hz, 50 Hz

(b)

8.00 m,

1.60 m

A heavy string is tied at one end to a movable support and to a light thread at the other end as shown in figure. The thread goes over a fixed pulley and supports a weight to produce a tension. The lowest frequency with which the heavy string resonates is 120 Hz. If the movable support is pushed to the right by 10 cm so that the joint is placed on the pulley, what will be the minimum frequency at which the heavy string can resonate?

,d Hkkjh Mksjh dk ,d fljk ,d Hkkjh ,oa xfr djus esa l{ke lgkjs ls cak/kh xbZ gS rFkk bldk nwljk fljk fp=kkuqlkj ,d /kkxs ls cka/kk x;k gSA /kkxk ,d fLFkj f?kjuh ls gksdj xqtj jgk gS rFkk blls ,d Hkkj ca/kk gqvk gS] ftlls ruko mRiUu gksrk gSA U;wure vko`fÙk ftlls Hkkjh Mksjh vuqukfnr gks ldrh gSA 120 gV~t Z gSA ;fn xfr'khy lgkjk nk;ha vksj 10 lseh [khpk tkrk gS] ftlls laf/k f?kuuh ij vk tkrh gS] og U;wure vko`fÙk fdruh gksxh] ftlls fd Hkkjh Mksjh vuqukfnr gks lds\

57.

240 Hz.


Page # 23

Wave Motion and String Waves

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