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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 51. |
When a wave pulse travelling in a string is reflected from rigid wall to which string is tied as shown in figure. For this situation two statements are given below. (1) The reflected pulse will be in same orientation of incident pulse due to a phase change of `pi` radians (2) During reflection the wall exert a force on string in upward directon for the above given two statements choose the correct option given belowA. Only `(1)` is trueB. Only `(2)` is trueC. Both are trueD. Both are wrong |
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Answer» Correct Answer - D Reflected pulse will be inverted a it is reflected by a denser medium. The well exerts force in downward direction. |
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| 52. |
A wire of density `9 gm//cm^(3)` is stretched between two clamps `1.00 m` aprt while subjected to an extension of `0.05 cm`. The lowest frequency of transverse vibration in the wire isA. `35 Hz`B. `45 Hz`C. `75 Hz`D. `90 Hz` |
| Answer» Correct Answer - A | |
| 53. |
Figure shows a string of linear mass density `1.0 g//cm` on which a wave pulse is travelling. Find the time taken (in milli second) by the pulse in travelling through a distance of `60 cm` on the string. Take `g = 10 m//s^(2)`. |
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Answer» Correct Answer - 30 `V = sqrt((T)/(mu)) = 20 m//s` `s = Vt` `t = 0.03` sec |
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| 54. |
A certain transverse sinusoidal wave of wavelength 20 cm is moving in the positive x direction. The transverse velocity of the particle at x=0 as a function of time is shown. The amplitude of the motion is: A. `(5)/(pi)cm`B. `(pi)/(2)cm`C. `(10)/(pi)cm`D. `2pi cm` |
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Answer» Correct Answer - c `V_(max)=Aomega=5 implies A(2pi)/(4)=5` `implies A=(10)/(pi)cm` |
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| 55. |
A certain transvers sinusoidal wave of wavelength `20 cm` is moving in the positive `X` direction. The transverse velocitt of the particle at `x = 0` as a function of time is shonwn. The amplitude of the motion is equal to `(x)/(pi)` `("in " cm)`. Find `x` : |
| Answer» Correct Answer - 10 | |
| 56. |
A transverse sinusoidal wave of amplitude `a`, wavelength `lambda` and frequency `f` is travelling on a stretched string. The maximum speed of any point in the string is `v//10`, where `v` is the speed of propagation of the wave. If `a = 10^(-3)m` and `v = 10ms^(-1)`, then `lambda` and `f` are given byA. `lambda=2pixx10^(-2)m`B. `lambda=10^(-3)m`C. `f 10^(3)//(2pi)Hz`D. `f=10^(4) Hz` |
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Answer» Correct Answer - a., c., d. `v_(max)=aomega=a(2pif)` Given that `a(2pif)=(v)/(10)=(10)/(10)=1` `f=(1)/(2pia)=(10^(3)/(2pi)Hz` As `v=flambda` or `lambda=(v)/(f)=(10)/(10^(3)//(2pi)=2pixx10^(2)m` |
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| 57. |
The equetion of a progressive wave travelling along a string is given by `y=10 sinpi(0.01x-2.00t)` where x and y are in centimetres and t in seconds. Find the (a) velocity of a particle at x=2 m and `t=5//6` s. (b) acceleration of a particle at x=1 m and `t=1//4` s. also find the velocity amplitude and acceleration amplitude for the wave. |
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Answer» The displacement equation for a particle is given by `y=10sinpi(0.01x-2.00t) cm` (a). Particle velocity is given by `(dy)/(dt)=-20.0pi cos(0.01x-2.00t) cm//s` putting `x=200 cm` and `t=5//6 s`, we get `(dy)/(dt)=-20.0pi cospi(2-(5)/(3))cm//s` `=-10.0pi cm//s` Also.the velocity amplitude `(dy)/(dt)]_(max) =20.0 pi cm//s` `(corresponding to`|cos pi(0.01x-2.00 t)|=1)` (b). Differentiating eq. (ii) w.r.t. time `t`, we get the particle aceleration as `(d^(2)y)/(dt^(2))=-40.0pi^(2) sinpi(0.01pi-2.00t) cm//s^(2)` Putting `x=100 cm` and `t =1//4 s`, we get `(d^(2)y)/(dt^(2))=-40.0pi^(2) sinpi(1-(1)/(2))cm//s^(2)` `=-40.0pi^(2) cm//s^(2)` Also, the acceleration amplitude `(d^(2)y)/(dt^(2)]_(max)=40.0pi^(2) cm//s^(2)` `(corresponding to`|sinpi(0.01x-2.00t)|=1)` |
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| 58. |
A wave travelling along X-axis is given by `y=2(mm) sin (3t-6x+pi//4)` where x is in centimetres and t in second. Write the phases and, hence, the find the phase difference between them at t=0 for two points on X-axis, `x=x_(1)=pi//3` cm and `x=x_(2)=pi//2` cm. |
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Answer» At `t=0` `y=sin(-6x+pi//4)` phase at `x_(1),phi=(-6xxpi/(3)+pi/(4))=(7)/(4)pi` and that at `x_(2) is (-6xxpi/(2)+pi/(4))=(11)/(4)pi` the phase difference is `|(7)/(4)pi-(11)/(4)pi|=pi`, the dusturbation at the two point are out of phase. |
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| 59. |
Four pieces of string of length L are joined end to end to make a long string of length 4L. The linear mass density of the strings are `mu,4mu,9mu and 16mu`, respectively. One end of the combined string is tied to a fixed support and a transverse wave has been generated at the other end having frequency `f` (ignore any reflection and absorption). string has been stretched under a tension `F`. Find the ratio of wavelength of the wave on four string, starting from right hand side.A. `12:6:4:3`B. `4:3:2:1`C. `3:4:6:12`D. `1:2:3:4` |
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Answer» Correct Answer - c Frequency of wave is same on all the four string. So, `lambda_(1)=(v_(1))/(f), lambda_(2)=(v_(2))/(f)` `lambda_(3)=(v_(3))/(f) lambda_(4)=(v_(4))/(f)` `lambda_(4), lambda_(3), lambda_(2), lambda_(1)=(1)/(40:(1)/(3):(1)/(2):1=3:4:6:12` |
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| 60. |
Four pieces of string of length L are joined end to end to make a long string of length 4L. The linear mass density of the strings are `mu,4mu,9mu and 16mu`, respectively. One end of the combined string is tied to a fixed support and a transverse wave has been generated at the other end having frequency `f` (ignore any reflection and absorption). string has been stretched under a tension `F`. Find the time taken by wave to reach from source end to fixed end.A. `(25)/(12)xx(L)/sqrt(F//mu)`B. `(10L)/sqrt(F//mu)`C. `(4L)/sqrt(F//mu)`D. `(L)/sqrt(F//mu)` |
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Answer» Correct Answer - b `v_(1)=sqrt((F)/(mu))` `v_(2)=sqrt((F)/(4mu))=(1)/(2)sqrt((F)/(mu))` `v_(3)=sqrt((F)/(9mu))=(1)/(3)sqrt((F)/(mu))` `v_(4)=sqrt((F)/(16mu))=(1)/(4)sqrt((F)/(mu)` Total time taken `=(L)/(v_(1)+(L)/(v_(2)+(L)/(v_(3)+(L)/(v_(4)=(10L)/(sqrt(F//mu))` |
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| 61. |
At t=0,a transverse wave pulse travelling in the positive x direction with a speed of `2 m//s` in a wire is described by the function `y=6//x^(2)` given that `x!=0`. Transverse velocity of a particle at x=2 m and t= 2 s isA. `3 m//s`B. `-3 m//s`C. `8 m//s`D. `-8 m//s` |
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Answer» Correct Answer - b `y(x,t)=(6)/((x-2t)^(2))` `implies v_(p)=(dely)/(delt)=(24)/((x-2t)^(3))` `v_(p)[x=2,t=2]=(24)/(-2^(3))=-3 m//s` |
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| 62. |
Staement I: pressure and density change do not occure in a transverse stationary wave. Statement II: the average distance between any two particles of the wave remains the same.A. Statement I is true, statement II is true and statement II is the correct explaination for statement I.B. Statement I is true, statement II is true and statement II is NOT the correct explaination for statement I.C. Statement I is true, statement II is false.D. Statement I is false, statement II is true. |
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Answer» Correct Answer - b in a travsverse vibration, the mean distance between the successive vibrating particle remains constant. Only crests and throughs are formed. |
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| 63. |
For a travelling harmonic wave `y=2.0 cos(10t-0.0080x+0.35)`, where x and y are in centimetres and t in seconds. What is the phase difference between oscillatory motion of two points separated by a distance of (a) 4cm (b) 0.5 cm (c ) `lambda//2` (d) `3lambda//4` |
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Answer» Given `y=2.0 cos (10t-0.0080x+0.35)`. The standard equation of travelling harmonic wave can be writtien as `y =A xcos(omegat-kx+phi)`. on comparing two equation, we have `omega =10 red//s` and `k=0.0080 cm^(-1)=0.80 m^(-1)` `lambda=(2pi)/(k)=(2pi)/(0.80)m` phase dufference `Deltaphi=(2pi)/(lambda)xxDeltax` (i). when `DeltaX=4 M,Delta phi=(2pi)/((2pi//0.80))xx4=3.2 rad` (ii) when `Deltax=0.5 m, Delta `phi=(2pi)/((2pi/0.80))xx0.5=0.40` rad (iii) when `Deltax=(pi)/(2), Delta phi=(2pi)/(lambda)xx(lambda)/(2)=pi rad` (iv). when `Deltax=(3lambda)/(4), Delta phi=(2pi)/(lambda)xx(3lambda)/(4)=(3pi)/(2) rad` |
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| 64. |
A transverse wave travelling on a taut string is represented by: `Y=0.01 sin 2 pi(10t-x)` Y and x are in meters and t in seconds. Then,A. The speed of the wave is `10m//s`.B. closet points on the string which differ in phase by `60^(@)` are `(1//6)m` apartC. maximum particle velocity is `pi//4 m//s`D. the phase of a certain point on the string changes by `120^(@)` in `(1//20)` seconds. |
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Answer» Correct Answer - a., b. `y=0.01 sin [2pi(10t-x)]` `impliesy=0.01sin[20pit-2pix]` `omega=20pi, K=2pi` Wave velocity :`v=(omega)/(k)=(20pi)/(2pi)=10m//s` `Deltaphi=K Deltax`, given `Deltaphi=60^(@)=(pi)/(3)` `(pi)/(3)=2pi DeltaximpliesDeltax=(1)/(6)m` `v_(p-max)=omegaA=20pixx0.01=(pi)/(5)m//s` `Deltaphi_(t)omegaDeltat=20pixx(1)/(20)=pi` so in time `(1)/(20)s`, phase changes by and not `120^(@)` |
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| 65. |
The wave function for a travelling wave on a taut string is `y(x,t)=(0.350 m)sin(10pit-3pix+pi//4)`. (SI units) (a) what is the speed and direction of travel of the wave ? (b) what if the verticaly position of an element of the string at `t=0, x=0.100 m?` (c ) what is the wavelength and frequency of the wave? (d) waht is the maximum transverse speed of an element of the string? |
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Answer» `Y(x,t) = (0.350m) sin (10pit - 3pi x + (pi)/(4))` comparing with equation , `Y = A sin (omegat - kx + phi) , omega = 10 pi , k = 3pi, f = (pi)/(4)` (a) speed `= (omega)/(k) = (10)/(3) = 3.33 m//sec` and along `+ve x` axis (b) `y(0.1, 0) = 0.35 sin (10 pi xx O - 3 pi (0.1) + (pi)/(4)) = 0.35 sin [(pi)/(4) - (3pi)/(10)] = -5.48 cm` (c) `k = (2pi)/(lambda) = 3 pi rArr lambda = (2)/(3) cm , = 0.67 cm` and `f = (v)/(lambda) = (10//3)/(2//3) = 5 Hz`. |
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| 66. |
For the wave shown in figure, the equation for the wave, travelling along `+x` acis with velocity `350 ms^(-1)` when its position at `t = 0` is as shown A. `0.05 sin ((314)/(4)x - 27475 t)`B. `0.05 sin ((379)/(5)x - 27475 t)`C. `1 sin ((314)/(4)x - 27475 t)`D. `0.05 sin ((289)/(5)x + 27475 t)` |
| Answer» Correct Answer - A | |
| 67. |
The wave-function for a certain standing wave on a string fixed at both ends is `y(x,t) = 0.5 sin (0.025pix) cos500t` where `x` and `y` are in centimeters and `t` is seconds. The shortest possible length of the string is :A. `126 cm`B. `160 cm`C. `40 cm`D. `80 cm` |
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Answer» Correct Answer - C `K = 0.025 pi = (2pi)/(lambda) lambda = (2cm)/(0.025)` Required length `= (lambda)/(2) = (1)/(0.025) = 40 cm` |
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| 68. |
Two sinusoidal waves of the same frequency travel in the same direction along a string. If `A_(1) =m 3.0 cm, A_(2) = 4.0 cm, phi_(1) = 0`, and `phi_(2) = pi//2 rad`, what is the amplitude of the resultant wave ? |
| Answer» Resultant amplitude `= sqrt(3^(2) + 4^(2) + 2 xx 3 xx 4 xx cos 90^(@)) = 5 cm`. | |
| 69. |
The displacement vs time graph for two waves A and B which travel along the same string arre shown in the figure. Their intensity ratio `I_(A)|I_(B)` is A. `9/4`B. 1C. `81/16`D. `3/2` |
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Answer» Correct Answer - b `(I_(1))/(I_(2))=(a_(1)^(2)f_(1)^(2))/(a_(2)^(2)f_(2)^(2))=((3)^(2)(8)^(2))/((2)^(2)(12)^(2))=1` |
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| 70. |
Two waves of equal frequency `f` and velocity `v` travel in opposite directions along the same path. The waves have amplitudes `A` and `3 A` . Then:A. the amplitude of the resulting wave varies with position between maxima of amplitude `4A` and minima of zero amplitudeB. the distance between a maxima and adjacent minima of amplitude is `(v)/(2f)`C. at a point on the path the average displacement is zero `rArr` Average displacement of medium particle of any point is zero.D. the position of a maxima or minima of amplitude does not change with time |
| Answer» Correct Answer - C::D | |
| 71. |
The streight shown in figure is driven at a frequency of `5.00 Hz`. The amplitude of the motion is `12.0 cm`, and the wave speed is `20.0 m//s`. Furthmore, the wave is such that `y = 0` and `t = 0`. Determine (a) the angular frequency and (b) wave number for this wave (c) Write an expression for the wave funcation. Calculate. (d) the maximum transverse speed and (e) the maximum transverse acceleration of a point on the string. |
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Answer» Correct Answer - (a) `10 pi rad//s` , (b) `pi//2 pi rad//s` , (c) `y = (12 xx 10^(-2)m) sin ((pi)/(2)x - 10 pit)` , (d) `(6)/(5) pi m//s` (e) `12 pi^(2) m//s^(2)` (a) `omega = 2pif = 10 pi` Red/sec `(b) lambda xx 5 = 20 , lambda = 4 m , K = (2pi)/(lambda) = (pi)/(2) rad//m` (c) ` y = 12 xx 10^(-2) sin (omegat - kx + phi) = 12 xx 10^(2) sin (10 pit - (pi)/(2) + phi)` At `t = 0 , x = 0 , y = 0` `(delx)/(delx) = 12 xx 10^(-2) (-(pi)/(2)) cos "(100 pit - (pi)/(2)x+phi)` At `t = 0 , x = 0` `(dely)/(delx) = -12 xx 10^(-2) ((pi)/(2)) cos phi` `(dely)/(delx)` should be positive `:. phi = pi` `y = 12 xx 10^(-2) sin (10 pit - (pi)/(2) + pi)= 12 xx 10^(-2) sin ((pi)/(2)xx-10 pit)` At `t = 0 , x = 0` `(dely)/(delx) = -12 xx 10^(-2) ((pi)/(2)) cos phi` `(dely)/(delx)` should be positive `y = 12 xx 10^(-2) sin (10pit - (pi)/(2)x + pi) = 12 xx 10^(-2) sin ((pi)/(2) x - 10 pit)` (d) `V_(max) = Aomega = 12 xx 10^(-2) xx 10 pi xx 10 pi = 12pi^(2)` (e) `A_("max") = Aomega^(2)` `= 12 xx 10^(-2) xx 10pi xx 10pi = 12 pi^(2)` |
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| 72. |
For plane waves in the air of frequency 1000 Hz and the displace amplitude `2xx10^(-8)m`, deduc (i) the velocity amplitude, and (ii) the intensity. (Take p= 1.3 kg//m^(3),c=340 m//s)` |
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Answer» `vmax("velocity amplitude")=aomega` `:.v_(max)=2xx10^(-8)xx(2pixx1000)=1.3xx10^(-4) m//s` `I=(1)/(2)pa^(2)omega^(2)c=(1)/(2)pv^(2)_(max)c` `implies I=(1)/(2)xx1.3(1.3xx10^(-4))^(2)xx340=3.7xx10^(-6) W//m^(2)` |
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| 73. |
A transverse wave on a string has an amplitude of 0.2 m and a frequency of 175 Hz. Consider a particle of the string at x=0. it begins with a displacement y=0, at t=0, according to equation `y=0.2 sin (kx+-omegat)`. How much time passed between the first two instant when this particle has a displacement of y=0.1 m?A. 1.9 msB. 3.9 msC. 2.4 msD. 0.5 ms |
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Answer» Correct Answer - a `y=(0.2 m) sin [kx+-omegat]` for `x=0, y=0.1 m` `0.1=0.2 sin(omegat)` `impliesomegat=pi//6 or 5pi//6` So, `t_(2)=5pi//6omega` and `T_(1)=pi//6omega` `t_(2)-t_(1)=2pi//3omega=1//3f=1.9 ms` |
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| 74. |
A water surfer is moving at a speed of `15 m//s`. When he is surfing in the direction of wave, he swing upwards every 0.8 s because of wave crests. While surfing in opposite direction to that of wave motion, he swings upwards every 0.6 s. determine the wavelength of transverse component of the water wave.A. 15 mB. 10.3 mC. 21.6 mD. information insufficient |
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Answer» Correct Answer - b Let the speed of wave be v, for crossing one wave crests to the other while travelling in the same direction, the surfing speed has to be greater than speed of the wave, i.e., `vlt15 m//s`. Let wavelength of wave be `lambda` m while surfing in the same direction, `lambda=(15-v)xx0.8` while surfing in the direction opposite to the wave motion, `lambda=(15+v)xx0.6` `(15-v)0.8=(15+v)0.6` `v=15//7 m//s=2.143 m//s` so, `lambda=(15-2.143)xx0.8=10.3 m` |
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| 75. |
At a moment in a progressive wave , the phase of a particle executing S. H. M `(pi)/(3)`. Then the phase of the particle 15 cm ahead and at the time `(T)/(2)` will be , if the wavelength is 60 cm |
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Answer» Correct Answer - `(5pi)/(6)` |
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| 76. |
A rope is attached at one end to a fixed vertical pole . It is stretched horizontal with a fixed value of tension `T` . Suppose at `t=0`, a pulse is generated by moving the free end of the rope up and down once with your hand. The pulse arrives at the pole at instant `t`. Ignoring the effect of gravity, answer the following quetions. A.. If you move your hand up and down once but to a greater distance and in the same amount of time.A. Time taken for the pulse to reach the pole will increase and it will be doubledB. Time taken for the pulse to reach the pole will deacrease and it will become halfC. Time taken for the pulse to reach the pole will not changeD. Cannot change |
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Answer» Correct Answer - c Time taken to reach other end is independent of frequency and amplitude `v=sqrt((T)/(m))` As `m` increases, velocity decreases. So time taken will be more or will increase. As `T` increase, velocity also increases. So time taken will be less or it will decreases. |
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| 77. |
A stretched rope having linear mass density `5xx10^(-2)kg//m` is under a tension of `80 N`. the power that has to be supoplied to the rope to generate harmonic waves at a frequency of `60 Hz` and an amplitude of `(2sqrt2)/(15pi)m` isA. 215 WB. 251 wC. 512 wD. 521 w |
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Answer» Correct Answer - c `P=(1)/(2)muomega^(2)A^(2)v` when `v=sqrt((T)/(mu))` |
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| 78. |
A stretched wire emits a fundamental note of frequency 256Hz. Keeping the stretching force constant and reducing the length of the wire by 10cm, frequency becomes 320Hz. Calculate original length of the wire. |
| Answer» Correct Answer - `50 cm` | |
| 79. |
A rope is attached at one end to a fixed vertical pole . It is stretched horizontal with a fixed value of tension `T` . Suppose at `t=0`, a pulse is generated by moving the free end of the rope up and down once with your hand. The pulse arrives at the pole at instant `t`. Ignoring the effect of gravity, answer the following quetions. If you move your hand up and down once but to a greater distance and in the same amount of time.A. Time taken for the pulse to reach the pole will increaseB. Time taken for the pulse to reach the pole will not changeC. Time taken for the pulse to reach the pole will deacreaseD. Time taken for the pulse to reach the pole may increase or decrease |
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Answer» Correct Answer - b Time taken to reach other end is independent of frequency and amplitude `v=sqrt((T)/(m))` As `m` increases, velocity decreases. So time taken will be more or will increase. As `T` increase, velocity also increases. So time taken will be less or it will decreases. |
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| 80. |
In a stationary waveA. all the particles of the medium vibrate eithr in phase or in oppsite phaseB. all the antinodes vibrate in opposite phaseC. all the particles between consecutive nodes vibrates in phaseD. all the particle between consecutive nodes vibrates in phase |
| Answer» Correct Answer - A::C::D | |
| 81. |
A plane progressive wave is given by `x=(40 cm) cos(50pit-0.02piy)` where `y` is in cm and `t` in `s`. The particle velocity at `y=25m` in time `t=(1)/(100)s` will be `10pisqrt(n) m//s`. What is the value of `n`. |
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Answer» Given that `x=40 cos (50pit-0.02piy)` :. Particle velocity `v_(p)=(dx)/(dt)=(40xx50pi){-sin(50pit-0.02piy)]` putting `x=25` and `t=(1)/(200)s`. `v_(p)=-(2000pi cm//s)sin[50pi((1)/(200))-0.02pi(25)]` `=10pisqrt(2) m//s` |
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| 82. |
Three progressive waves `A,B` and `C` are shown in figure. With respect to wave `A` .A. The wave C is ahead by a phase angle of `pi//2` and the wave B lags behicnd by a phase angle of `pi//2`B. The wave C lags behind by a pahse angle of `pi//2` and the wave B is ahead by a phase angle of `pi//2`C. The wave C is ahead by a phase angle of `pi` and the wave B lags behind by a phase angle of `pi`D. The wave C lags behind by a phase angle of `pi` and the wave B ahead by a phase angle of `pi` |
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Answer» Correct Answer - b |
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| 83. |
Give two difference between progressive wave and stationary wave. |
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Answer» Correct Answer - 1. In a progressive wave, disturbance travels for work in the medium; while in a stationary wave, the disturbance does not move forward or backward. 2. In a progressive wave, there is no transmission of energy while in a stationary wave, the is not transimission of energy. |
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| 84. |
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^-1` 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 ? |
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Answer» `L=40 cm, mass=10 g` mass per unit length `mu=(10)/(40)=(1)/(4)(g//cm)` spring constant `k=160 N//m` deflection, `x=1 cm=0.01 m` tension in the string: ` T=kx=160xx0.01=1.6 N` `=16xx10^(4)"dyne"` wave velocity is given by `v=sqrt(((t)/(mu)))=sqrt((((16xx10^(4)))/((1)/(4))))=800 cm//s` Time taken by the pulse to reach the spring `t=(40)/(800)=(1)/(20)=0.05 s=5xx10^(-2)s`. |
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| 85. |
The equation of a progressive wave is `y=0.02sin2pi[(t)/(0.01)-(x)/(0.30)]` here x and y are in metres and t is in seconds. The velocity of propagation of the wave isA. `300 m//s`B. `30 m//s`C. `400 m//s`D. `40 m//s` |
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Answer» Correct Answer - b `omega=(2pi)/(0.01)` and `k=(2pi)/(0.30)` `v=(omega)/(k)=(2pi)/(0.01)xx(0.30)/(2pi)=30 m//s` |
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| 86. |
A progressive wave is given by `y=3 sin 2pi [(t//0.04)-(x//0.01)]`where x, y are in cm and t in s. the frequency of wave and maximum accelration will be:A. `100 Hz,4.7xx10^(3) m//s^(2)`B. `50 Hz, 7.5xx10^(3) m//s^(2)`C. `25 Hz, 4.7xx10^(4) m//s^(2)`D. `25 Hz, 7.5xx10^(4) m//s^(2)` |
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Answer» Correct Answer - d We know `(omega)=2pif=(2pi)/(0.04)` `rArr f=285 Hz` Differented `y` w.r.t. twice, we have `y'(-3xx(4pi^(2)))/((0.004)^(2)) sin(2pi[((t)/(0.04))-((x)/(0.01))])` For maximum acceleration `y_(0)'(-3xx(4pi^(2)))/((0.004)^(2))=7.5xx10^(4) m//s^(2)` |
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| 87. |
At t=0, the shape of a travelling pulse is given by `y(x,0)=(4xx10^(-3))/(8-(x)^(-2)` where x and y are in metres. The wave function for the travelling pulse if the velocity of propagation is `5 m//s` in the x direction is given byA. `y(x,t)=(4xx10^(-3))/(8-(x^(2)-5t))`B. `y(x,t)=(4xx10^(-3))/(8-(x-5t)^(2))C. `y(x,t)=(4xx10^(-3))/(8-(x+5t)^(2))`D. `y(x,t)=(4xx10^(-3))/(8-(x^(2)+5t))` |
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Answer» Correct Answer - b `y(x,t)=f(x-vt)` `y=(x,0)=(4xx10^(-3))/(8-x^(2))` For a travelling wave in the `x-`direction `y(x,t)=(4xx10^(-3))/(8-(x-5x)^(2))` |
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| 88. |
A sinusoidal wave on a string is described by the wave function where x and y are in metres and t is in seconds. The mass per unit length of this string is `12.0 g//m`. Determine (a) the speed of the wave, (b) the wavelength, (c ) the frenquecy and (d) the power transmitted to the wave. |
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Answer» Comparing the given wave function, `y=(0.15 m0sin (0.80x-50r)` with the general wave function, `y=A sin (ks-omegat)` We have k`=0.80 rad//m` and `omega=50 rad//s and A=0.15 m` `(a) the wave speed is then `v=flambda=(omega)/(k)=(50.0 rad//s)/(0.80 rad//m)=62.5 m//s` (b) The wavelength is lambda=(2pi)/(k)=(2pi rad)/(0.80 rad/m)=7.85 m` (c ) The frequency is `f (omega)/(2pi) (50 rad//s)/(2pi and) 7.96 Hz` (d) The wave carries power `pmu(1)/(2)muomega^(2)A^(2)v` `mu(1)/(2)(0.0120 kg//m)(50.0 s^(-1))^(2)(0.150 m)^(2)(62.5 m//s)` `mu21.1 W` |
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| 89. |
Fore a transverse wave on a string, the string displacement is describrd by `y(x,t)=f(x-at)` where f represent a function and a is a negative constant. Then which of the following is//are correct statement(S)?A. the shape of the string at time `t=0` is given by `f(x)`B. the shape of wave form does not change as it moves along the stringC. waveform moves in +ve x-directionD. the speed of waveform is a |
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Answer» Correct Answer - a.,b.,d. `y(x,0)=f(x)`. So, shape of string at `t=0` is given by `y=f(x)` As velocity of wave `dx//dt=+a`, is constant, so shape of string does not change, or we can say `(x-at)` is constant. Thus, the shape of string remains the same. As `a` is `-ve` and constant, so `dx//dt =-ve` and hence, wave is moving along `-ve x-`direction. speed `=-a` and not a as speed cannot `be-ve` |
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| 90. |
If a wave is going from one medium to another, thenA. frequencyB. wavelengthC. velocityD. amplitude |
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Answer» Correct Answer - b., c., d. Frequency is the property of source while velocity is the property of medium and wavelength is the property of both medium and source. So, wavelength and velocity of wave must change as the medium change, while frequency remains same. On the amplitude (and hence intency) can decrease as the medium change. Amplitiude will either decrease or remain the same but it can never increase due to change in medium (assuming no external source is providing energy). |
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| 91. |
statement I: The intensity of a plane progressive wave does not change with change in distance from the source. Statement II: The wavefronts associated with a plane progressive wave are planar.A. Statement I is true, statement II is true and statement II is the correct explaination for statement I.B. Statement I is true, statement II is true and statement II is NOT the correct explaination for statement I.C. Statement I is true, statement II is false.D. Statement I is false, statement II is true. |
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Answer» Correct Answer - a Since the wavefront are plane, the amount of energy passing per unit time per unit area remains same. |
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| 92. |
Mark the correct option(s) out of the following:A. Mechanical waves can be transverse in liquids.B. in some medium, the speed of a longitudinal mechanical wave is greater than the speed of transverse mechanical wave.C. transverse waves ate possible in bulk of a liquid.D. Non-mechanical waves are transverse in nature. |
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Answer» Correct Answer - a., b., d. Mechanical waves can be transverse on a liquid surface and this is possible only because of surfacing tension. solid, `V_("longitudinal")gtv_("transverse")` Transverse waves are possible only on the surface of a liquide because they require the peroperty of rigidity. all non-mechanical waves found till now are transverse in nature. |
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| 93. |
A wave moves ar a constant speed along a stretched string.Mark the incorrect statement out of the following:A. particle speed is constant and equal to the wave speedB. particle speed is independent of amplitude of the periodic mmotion of the source.C. particle speed is independent of frequency of periodic motion of the sourceD. particle speed is dependent on tension and linear mass density the string |
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Answer» Correct Answer - a.,b.,c.,d. `v_(p)=vxx dely//delx "for" y=A sin (omegat-kx)`, `v_(p)=kvxxA cos(omegat-kx)=omega A cos(omegat-kx)` i.e., it is varying Also, `v_(p) prop omega` and `v_(p) omega A` |
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| 94. |
A sinusoidal wave travelling in the positive direction on a stretched string has amplitude `2.0 cm`, wavelength `1.0 m` and velocity `5.0 m//s`. At `x = 0` and ` t= 0` it is given that `y = 0` and `(dely)/(delt) lt 0`. Find the wave function `y (x, t)`.A. `Y(x,t)=(0.02 m)sin[(2pi m^(-1))x+(10pis^(-1))t]m`B. `y(x,t)=(0.02 m)cos(10pis^(-1))t+(2pi m^(-1))xm`C. `y(x,t0=(0.02 m)sin[(2pi m^(-1))x-(10pi s^(-1))t]m`D. `y(x,t)=(0.02m)sin[(pim^(-1))x+(pis^(-1))t]m` |
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Answer» Correct Answer - c We start with a general form for a rightward moving wave, `y(x,t)=A sin(kx-omegat+phi)` The amplitude given is `A=2.0 cm=0.02 m`. The wavelength is given as, `lambda=1.0 m` wave number `=k=2pi//lambda=2pi m^(-1)` Angular frequency, `omega =vk =10pi rad //s` `y(x,t)=(0.02) sin [2pi(x-5.0 t)+phi]` we are told that for `x=0, t=0`, `y=0` and `(dely)/(delt)lt0` i.e., `0.02 sin phi =0` `(as y=0)` and `-0.2 pi cos philt0` From these condition, we may conclude that `phi=2npi` where `n=0,2,4,6,......` Therefore, `y(x,t)=(0.02 m) sin [(2pi m^(-1))x-(10pi s^(-1))t] m` |
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| 95. |
A sinusoidal wave with amplitude `y_(m)` is travellling with speed `V` on a stgring with linear density `rho`. The angular frequency of the wave is `omega`. The following conclusions are down. Mark the one which is correct.A. doubling the frequency doubles the rate at which energy is carried along the stringB. if the amplitude were doubled, the rate at which energy is carried would be halvedC. if the amplitude were doubled, the rate at which energy is carried would be doubledD. the rate at which energy is carried is directly proportional to the velocity of the wave. |
| Answer» Correct Answer - D | |
| 96. |
A wave moving with constant speed on a uniform string passes the point `x=0` with amplitude `A_0`, angular frequency `omega_0` and average rate of energy transfer `P_0`. As the wave travels down the string it gradually loses energy and at the point `x=l`, the average rate of energy transfer becomes`P_0`/2. At the point `x=l`. Angular frequency and amplitude are respectively:A. `omega_(0)` and `A_(0)//sqrt(2)`B. `omega_(0)//sqrt(2)` and `A_(0)`C. less than `omega_(0)` and `A_(0)`D. `omega_(0)//sqrt(2)` and `A_(0)//sqrt(2)` |
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Answer» Correct Answer - A `p_(0) = A_(0^(2)) omega_(0^(2)) muv` `(p_(0))/(2) = A^(2)omega^(2)muv` `:. 2 = (A_(0^(2)) omega_(0^(2)))/(A^(2)omega^(2))` `:.` As, `omega = omega_(0)` (frequency remains then same) `:. A = (A_(0))/(sqrt(2))` |
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| 97. |
Show that `(a)y=(x+vt)^(2),(b)y=(x+t)^(2),(c )y=(x-vt)^(2)`, and (d) y=2 sin xcos vt are each a solution of one dimensional wave equation but not (e) y=x^(2)-v^(2)t^(2)` and (f) `y=sin 2x cos vt.` |
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Answer» Differentiating expression (i) twice `w.r.t.t` we have, `(del^(2)y)/(delx^(2)=(2)` And differentiating expression (i) twice `w.r.t.x` we have, `(del^(2)y)/(delx^(2)=2v^(2)` clearly, `(del^(2)y)/(delt^(2))=v^(2)(del^(2)/delx^(2))` Thus expression (i) is a solution of the one dimensional wave equation. Similarly, treatment can be done for (ii), (iii) and (iv). note: the expression (iv) satisfies differential equation of wave, but it does not represent progressive wave. it represents stationary wave. (v). Differentiating expression (v) twicee `w.r.t.t` we have `(del^(2)y)/delt^(2)=-2v^(2)` clearly `(del^(2)y)/(delt^(2)!=v^(2)(del^(2)y)/(delx^(2)`, so the expression (v) is not a solution of the one dimensional wave equation. (vi) Differentiating expression (vi) twice `w.r.t.t, we have `(del^(2)y)/(delt6(2))=-v^(2) sin2xcos vt=-v^(2)y` `(del^(2)y)/(delx^(2)=-4y` clearly `(del^(2)y)/(delt^(2)!=v^(2)(del^(2)y)/(delx^(2)` and therfore the expression is not a solution of the one dimensional wave equation. |
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| 98. |
A wave moving with constant speed on a uniform string passes the point `x=0` with amplitude `A_0`, angular frequency `omega_0` and average rate of energy transfer `P_0`. As the wave travels down the string it gradually loses energy and at the point `x=l`, the average rate of energy transfer becomes`P_0`/2. At the point `x=l`. Angular frequency and amplitude are respectively:A. `omega_(0)"and "(A_(0))/(sqrt(2))`B. `omega_(0)/(sqrt(2))"and"A_(0)`C. less then `omega_(0)` and `A_(0)`D. `(omega_(0))/(sqrt(2))` and `(A_(0)0/(sqrt(2))` |
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Answer» Correct Answer - A |
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| 99. |
A wave travels out in all direction from a point source. Justify the expression `y=(a_(0)/r) sin k (r-vt)`, at a distance r from the source. Find the speed, periodicity and intensity of the wave.what are the dimensions of `a_(0)`? |
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Answer» Let `P` be the power of the source. Then `I=(P)/(4pir^(2)` `:. Iprop(1)/(r^(2)) but Ipropa^(2)` `:. Aprop(1)/(r )` The equation in the standard form is `y=a sin k (r-vt)` Therefore, `y=a_(0)/r sin k(r-vt)`, simply replacing a by `(a_(0))/(r)` where `a_(0)` is a constant. Comparing with the usual form `y=a sin (kr-omegat)` `omega=kv` or `f=kv//2pi` and `k=2pi//lambda or lambda=2pi//k` `:. c=flambda=(kv)/(2pi)xx(2pi)/(k)=v` and `T=1//f=2pi//kv` `I=(1)/(2)pa^(2)omega^(2)v=(1)/(2)p(a_(0)^(2))/(r^(2))k^(2)v^(2)v=(1)/(2)pa_(0)^(2)k^(2)v^(3)//r^(2)` |
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| 100. |
A wave moving with constant speed on a uniform string passes the point `x=0` with amplitude `A_0`, angular frequency `omega_0` and average rate of energy transfer `P_0`. As the wave travels down the string it gradually loses energy and at the point `x=l`, the average rate of energy transfer becomes`P_0`/2. At the point `x=l`. Angular frequency and amplitude are respectively:A. directly on the square of the wave amplitude and square of the wave frequencyB. directly on the square of the wave amplitude and square root of the wave frequencyC. directly on the wave frequency and square of the wave amplitudeD. directly on the wave frequency and square of the wave frequency |
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Answer» Correct Answer - A By definition |
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