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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.
| 101. |
Two wave of amplitude `A_(1)`, and `A_(2)` respectively and equal frequency travels towards same point. The amplitude of the resultant wave isA. `A_(1) + A_(2)`B. `A_(1) - A_(2)`C. between `A_(1) + A_(2)` and `A_(1) - A_(2)`D. Can not say |
| Answer» Correct Answer - C | |
| 102. |
Does the wave function `y=A_(0) cos^(2)(2pi f_(0)t-2pix//lambda_(0)` represent a wave? If yes, the determine its amplitude frequency, and wavelength. |
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Answer» Any wave function represents a wave if it satisfies the equation ltbRgt ` (del^(2)y)/(delx^(2))=(1)/(v^(2))(del^(2)y)/(delt^(2)` It can be easily proved that the given wave function satisfies the above equation. Furthermore, it can be easily put in the form `y=f(x-vt)`. thus, it represent a wave. the amplitude, frequency, and wavelength of the wave are `A_(0)//2,2f_(0)` and `lambda_(0)//2`, respectively. it is because, the given function can be reduced to `y=(A_(0))/(2)[1+cos{2(2pif_(0)t-(2pix)/(lambda_(0))}]` `(A_(0))/(2)[1+cos[2pi(2f_(0))t-(2pix)/(lambda_(0)//2)]]` |
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| 103. |
When a train of plane wave traverses a medium, individual particles execute periodic motion given by the equation `y 4sin(pi)/(2)(2t x/8)` Where the length are expressed in centimetres and time in seconds. Calculate the amplitute, wavelength, (a) the phase different for two positions of the same particle which are occupied at time interval 0.4 s apart and (b) the phase difference at any given instant of two particle 12 cm apart. |
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Answer» The equation of a wave motion is given by `y=A sin(2pi)/(lambda)(vt+x)`(i) Here, `y 4 sin(pi)/(2)(2t x/8)` This equation canbe written as `y 4sin(2pi)/(32)(16t x)` (ii) comparing `Eq`. (i)with `Eq`. (ii), we get Amplitude `A=4 cm,`wavelength `(lambda)=32 cm`, wave velocity `v=16 cm//s` here frequency is given as `f=(v)/(lambda)=(16)/(32)=(1)/(2)=0.5 Hz` (a). phase of a particle at instant `t_(1)` is given by `phi_(1) (pi)/(2)(2t_(1) x/8)` The phase at instant `t_(2)` is given by `phi_(2) (pi)/(2)(2t_(1) x/8)` The phase difference is given as `phi_(1)-phi_(2)=(pi)/(2)[(2t_(1)+x/8)-(2t_(2)+x/8)]` `pi(t_(1)-t_(2)) pi(0.4)` `(As t_(1)-t_(2)=0.4)` `=180xx0.4=72^@` `(pi rad=180^@)` (b). phase different at an instant between two particle with path different `(Delta)` is `phi=(2pi)/(lambda)xxDelta` `(2pi)/(32)xx12` `(As Delta=12 cm` `(3pi)(4)` |
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| 104. |
A wave pulse starts propagating in the +x-direction along a non-uniform wire of length 10 m with mass per unit length given by `mu=mu_(0)+az` and under a tension of 100 N. find the time taken by a pulse to travel from the lighter end (x=0) to the heavier end. `(mu_(0)=10^(2) kg//m and a=9xx10^(-3)kg//m^(2)). |
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Answer» The speed of the wave pulse `v=sqrt(T)/(mu)=sqrt((T)/(mu_(0)+ax)=(dx)/((dt)` `:. T=int_(0)^(t) dt=int_(0)^(L) sqrt((mu_(0)+ax)/(T) )dx=(2)/(3)(1)/(asqrt(T))[(mu_(0)+aL)^(3//2)-(mu_(0))^(3//2)]` substituting the value, we get `t=(2)/(3)xx(1)/(9xx10^(-3)xxsqrt(100))[(10^(-2)+9xx10^(-3)xx102))(3//2)-(10^(-2)^(3//2]` `=(2xx100)/(27)[(10)^(-3//2)-(10)(-3)]=0.227 s` |
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| 105. |
A circular loop of rope of length L rotates with uniform angular velocity `omega` about an axis through its centre on a horizontal smooth platform. Velocity of pulse (with resppect to rope ) produce due to slight radiul displacement is given by `(omegaL)/(2pi)` , if the motion of the pulse and rotation of the loop, both are in same direction then the velocity of the pulse w.r.t. to ground will be:A. `omegaL`B. `(omegaL)/(2pi)`C. `(omegaL)/(pi)`D. `(omegaL)//(4pi^(2))` |
| Answer» Correct Answer - C | |
| 106. |
A circular loop of rope of length L rotates with uniform angular velocity `omega` about an axis through its centre on a horizontal smooth platform. Velocity of pulse (with resppect to rope ) produce due to slight radiul displacement is given by A. `omegaL`B. `(omegaL)/(2pi)`C. `(omegaL)/(pi)`D. `(omegaL)/(4pi^(2))` |
| Answer» Correct Answer - B | |
| 107. |
A wire of `9.8xx10^(-3) kg/m` passes over a frictionless light pulley fixed on the top of a frictionless inclined plane which makes an angle of `30^(@)` with the horizontal. Masses m and M are tied at the two ends of wire such that m rests on the plane and M hangs freely vertically downwards. the entire system is in equilibrium and a transverse wave propagates along the wire with a velocities of `100 m//s`.A. `m=20 kg`B. `M=5 kg`C. `(m)/(M)=(1)/(2)`D. `(m)/(M)=2` |
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Answer» Correct Answer - a., d. `v=sqrt((T)/(mu))` For equilibrium `Mg=mg sin 30=T` `M=m//2` `100=sqrt((Mg)/(9.8xx10^(-3)))=sqrt((M(9.8))/(9.8xx10^(-3))` `100=sqrt(M(1000))` `M=10 kg and m=20 kg` |
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| 108. |
A travelling wave is given by `y(0.8)/((3x^(2)+24xt+48t^(2)+4))` where `x` and `y` are in metres and `t` is in seconds. Find the velocity in `m//s`. |
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Answer» `y=(0.8)/((3x^(2)+24xt+48t^(2)+4))=(0.8)/(3[x^(2)+8xt+16t^(2)]+4)` `=(0.8)/(3(x+4t)^(2)+4)` `:.x+4t=x+vt` `:.v=4 m//s` |
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| 109. |
A travelling wave is given by `y=(0.8)/((3x^(2)+24xt+48t^(2)+4))` where `x` and `y` are in metres and `t` is in seconds. Find the velocity in `m//s`. |
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Answer» `y=(0.8)/((3x^(2)+12xt+12t^(2)+4))` `=(0.8)/(3[x^(2)+4xt+(2t)^(2)]+4)` `=(0.8)/([4+3(x+2t)^(2)])` on solving, we get `x+2t=x+vt` `v=2 m//s` Amplitude ua `y_(max):` when `x+2t=0` therefore, `y_(max)=(0.8)/(4)=0.2m` |
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| 110. |
Figure shows two snapshots of medium particle within a time interval of `1/60` s. find the possible time periods of the wave |
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Answer» The wavelengt of the wave `lambda=(12-0)=(14-2)=12 units` therefore, in time `(1//60` s, the distance moved by the wave is `d=2+n(lambda)=lambda//60+nlambda,n "in" J` velocity of wave: `v=(d)/((1//60))=(lambda)/(6)+nlambda)60` `T=(lambda)/(v)` `=(lambda)/(((lambda)/(6)+nlambda)60)=(1)/(10(6n+1)` |
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| 111. |
The mathmaticaly form of three travelling waves are given by `Y_(1)=(2 cm) sin (3x-6t)` `Y_(2)=(3 cm) sin (4x-12t)` And `Y_(3)=94 cm) sin(5x-11t)` of these waves,A. Wave 1 has greatest wave speed and greatest maximum transverse string speedB. wave 2 has greatest wave speed and wave 1 has greatest maximum transverse string speedC. wave 3 has greatest wave speed and wave 1 has greatest maximum transverse string speedD. wave 2 has greatest wave speed and wave 3 has greatest maximum transverse string speed |
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Answer» Correct Answer - d For the wave, `y=A sin (kx-omegat)`,the wave speed is `omega//k` and the maximum transvers string speed is `Aomega`. |
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| 112. |
The shows two snapshots, each of a wave travelling along a particular string. The phase for the waves are given by (a) 4x-8t (b) 8x-16t. Which phase corresponds to which waves in |
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Answer» The standard equation of a plane progressive wave is `y=A sin (kx-omegat)`, the phase of the wave is `(kx-omegat)`, we get `omega =16 and k=8` `lambda_(2)=(2pi)/(k)=(2pi)/(8)=(pi)/(4)` clearly `lambda_(1)=2lambda_(2)` Therefore, snapshots `1` and `2` correspond to a and `b` respectly. |
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| 113. |
Two waves travelling in opposite directions produce a standing wave . The individual wave functions are given by `y_(1) = 4 sin ( 3x - 2 t)` and `y_(2) = 4 sin ( 3x + 2 t) cm` , where `x` and `y` are in cm |
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Answer» (a) When the two waves are summed. The result is a standing wave whose mathematical representation is given by Equation, with `A = 4.0 cm` and `k = 3.0 rad//cm,` `y = (2A sin kx)cos omegat = [(8.0 cm) sin 3.0 X] cos 2.0 t` Thus, the maximum displacement of a particel at the position `x = 2.3 cm` is `y_(max) = [(8.0 cm) sin 3.0x]_(x = 2.3 cm)` `= (8.0 m) sin (6.9 rad) = 4.6 cm` (b) Because `k = 2pi//lambda = 3.0 rad//cm`, we see that `lambda = 2pi//3cm`. Therefore, the antinodes are located at `x = n((pi)/(6.0)) cm (n = 1, 3, 5, ........)` and the nodes are located at `x = n(lambda)/(2-)((pi)/(3.0)) cm (n = 1, 2, 3, .........)` |
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| 114. |
Two waves, each having a frequency of `100 Hz` and a wavelength of `2 cm`, are travelling in the same direction on a string. What is the phase difference between the wayes (a) if the second wave was produced `10 m` sec later than the first one at the same place (b) if the two waves were produced at a distance `1 cm` behind in the second one? (c) if each of the wave has an amplitude of `2.0 mm`, what would be the amplitude of the resultant waves in part (a) and (b) ? |
| Answer» Correct Answer - (a) `2pi` , (b) `pi` , (c) `4.0 mm`, zero | |
| 115. |
What are (a) the lowest frequency (b) the scond lowest frequency and (c) the third lowest frequency for standing waves on a wire that is `10.0 m` long has a mass of `100 g` and is stretched under a tension of `25 N` which is fixed at both ends? |
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Answer» Correct Answer - (a) `2.5 Hz;` , (b) `5 Hz` , (c) `7.5 Hz`. `v = sqrt((T)/(mu)) = sqrt((25)/((100 xx 10^(3))/(10))) = 50` `:. V_(0) = (v)/(2L) = (50)/(2 xx 10) = 2.5 Hz :. V_(1) = 2v_(0) = 5 Hz` `v_(2) = 3v_(0) = 7.5 Hz`. |
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| 116. |
A nylon guiter string has a linear density of `7.20 g//m` and is under a tension of `150 N`. The fixed supports are distance `D = 90.0 cm` apart. The string is osillating in the stading wave pattern shown in figure. Calculate the (a) speed. (b) wavelength and (c) frequency of the travelling waves whose superposition gives this standing wave. ltbgt |
| Answer» Correct Answer - (a) `(250)/(sqrt(3))m//s;` , (b) `60.0 cm;` , (c) `(1250)/(3sqrt(3)) Hz` | |
| 117. |
A wave travelling in positive X-direction with `A=0.2m` has a velocity of `360m//sec` . If `lambda=60m` , then correct expression for the wave isA. `y=0.2sin[2pi(6t+(x)/(60))]`B. `y=0.2sin[pi(6t+(x)/(60))]`C. `y=0.2sin[2pi(6t-(x)/(60))]`D. `y=0.2sin[pi(6t-(x)/(60))]` |
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Answer» Correct Answer - c |
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| 118. |
A wave is represented by the equetion`y=7 sin(7pi t-0.04pix+(pi)/(3))` x is in metres and t is in seconds. The speed of the wave isA. `175 m//s`B. `49pi m//s`C. `49//pi m//s`D. `0.28 pi m//s` |
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Answer» Correct Answer - a standard equation `y=A sin (omegat -kx+phi_(0))` In a given equetion `omega=7pi, k=0.04pi` `v=(omega)/(k)=(7pi)/(0.04pi)=175 m//s` |
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| 119. |
A wave is represented by the equation `y=A sin314[(t)/(0.5s)-(x)/(100 m)]` The frequency is n and the wavelength is `lambda`.Then:A. `n=2 Hz`B. `n=100 Hz`C. `lambda=2m`D. `lambda=100 m` |
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Answer» Correct Answer - b., c. The equation has to be reduced to the form `y=A sin2pi((t)/(T)-(x)/(lambda))` `A sin314((t)/(0.5s)-(x)/(100 m))` `A sin 2pi((50t)/(0.5s)-(x50)/(100m))` `=A sin2pi((t)/(0.01s)-(x)/(2m))` `n=(1)/(T)=(1)/(0.01)=100Hz` and `lambda=2 m` |
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| 120. |
A wave pulse passing on a string with speed of `40cms^-1` in the negative x direction has its maximum at `x=0` at` t=0`. Where will this maximum be located at `t=5s?` |
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Answer» `v=40 cm//s` As velocity of a wave is constant, location of maximum after `5` s is given by `40xx5=200 cm` along the negative `x`-axis at `x=-2 m` |
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| 121. |
A pulse is started at a time `t=0` along the `+x`direction on a long, taut string. The shape of the pulse at `t=0` is given by function `f(x)` with here `f` and `x` are in centimeters. The linear mass density of the string is `50 g//m` and it is under a tension of `5N`. The transverse velocity of the particle at `x=13 cm` and `t=0.015 s` will beA. `-250 cm//s`B. `-500 cm//s`C. `500 cm//s`D. `-1000 cm//s` |
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Answer» Correct Answer - a Transverse velocity `=(dely)/(delt)att=0.015 s, vt=15 cm` As for `x=13 cm` `(vt-4)ltxltvt` Therefore, `(dely)/(delt)=-(v)/(4)=-250 cm//s` |
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| 122. |
A pulse is started at a time `t=0` along the `+x`direction on a long, taut string. The shape of the pulse at `t=0` is given by function `f(x)` with ` {((x-vt)/4+1,for,vt-4ltx,le,vt),(-(x-vt)+1,for,vt,ltxlt,vt+1):}` `0`, otherwise `{((x-vt)/4+1,for,vt-4ltx,le,vt),(-(x-vt)+1,for,vt,ltxlt,vt+1):}` `0`, otherwise here `f` and `x` are in centimeters. The linear mass density of the string is `50 g//m` and it is under a tension of `5N`. The verticle displacement of the particle of the string at `x=7 cm` and `t=0.01 s` will beA. `0.75 cm`B. `0.5 cm`C. `0.25 cm`D. `zero |
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Answer» Correct Answer - c `v=sqrt((T)/(mu))=10 m//s` solution of the wave equation that gives displacement of any piece of the string at any time `{((x-vt)/4+1,for,vt-4ltx,le,vt),(-(x-vt)+1,for,vt,ltxlt,vt+1):}` `0`, otherwise Using `v=1000 cm//s,t=0.01s` `vt=10 cm` as `(vt-4)lt(x=7 cm)ltvt` `y=(1)/(4)(7-10)+1+(1)/(4)cm=0.25cm` |
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| 123. |
The diagram below shows the propagation of a wave. Which points are in same phase? .A. F, GB. C and EC. B and GD. B and F |
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Answer» Correct Answer - d |
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| 124. |
Which of the following curves represents correctly the oscillation given by `y=y_0sin(omegat-phi)` where `0lt phi lt 90^@`? .A. AB. BC. CD. D |
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Answer» Correct Answer - d |
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| 125. |
As a wave propagates,A. the wave intensity remain constant for a plane waveB. the wave intensity decrease as the inverse of the distance from the source for a sphericalC. the wave ontensity decrease as the inverse square of the distance from for a spherical waveD. total intensity of the spherical wave over the spherical surface centered at the source remains constant at all times |
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Answer» Correct Answer - A::C::D |
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| 126. |
A plane wave propagates along positive x-direction in a homogeneous medium of density `p=200 kg//m^(3)`. Due to propagation of the wave medium particle oscillate. Space density of their oscillation energy is `E=0.16pi^(2) J//m^(3)` and maximum shear strain produced in the mendium is `phi_(0)=8pixx10^(-5)`. if at an instant, phase difference between two particles located at points `(1m,1m,1m)` and `(2m, 2m, 2m,)` is `Deltatheta=144^(@)`, assuming at `t=0` phase of particle at `x=0` to be zero, Equation of wave isA. `y pi 10^(-4)sinpipi2000t-0.8xpi`B. `y pi 10^(-4)sinpipi400t-0.8xpi`C. `y pi 10^(-4)sinpipi100t-8xpi`D. `y pi 10^(-4)sinpipi100t-2xpi` |
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Answer» Correct Answer - b Since, the wave is a plane travelling wave, intensity at every point will be the same, since, initial phase of particle at `x=0` is zero and the wave is travelling along positive `x-` deirection equation of the wave will be of the form `delta=a sin omega (t-(x)/(v))` ...`(i)` Let intensity of the wave be `I`, then space density of oscillation energy of medium particles will be equal to `E=(I)/(v)` But, `I=2pi^(2)n^(2)a^(2)p=0.16pi^(2)J//m^(3)` `a^(2)n^(2)=4xx10^(-4)` or, `an=0.02` shear strain of the medium is `phi=(d)/(dx)delta` Differentiating Eq. (i), `phi=-(aomega)/(v)cosomega(t-(x)/(v))` Modulus of shear strain `f` will be maximum when `cosomega(t-(x)/(v))=+-1` `:.`maximum shear srain `8pixx10^(-5)` `phi_(0)=(aomega)/(v)` but it is equal to `(aomega)/(v)=8pixx10^(-5)` where `omega=2pin` `an=4vxx10^(-5)` ....`(iii)` Solving Eqs. (ii) and (iii), `v=500 m//s` since, the wave is travelling along positive `x-`direction, there, phase difference is give by `Deltatheta=2pi(Deltax)/(lambda)` `Deltax=(x_(2)-x_(1))=(2-1)m=1m` `lambda=(2piDeltax)/(Deltatheta)=2.5 m` But `v=nlambda`,therefore, `n=(v)/(lambda)=200 Hz` substituting `n=200 Hz` in Eq. (ii), `a=1xx10^(-4) m` Angular frequency,`omega=2pi n=400 pi rad//s`. substituting all these values in Eq. (i), `Delta=10^(-4) sinpi(400t-0.8x)m` since, due to propagation of the wave, shear strain is produced in the medium, the wave is a plane transverse wave. |
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| 127. |
Which properties of a medium are responsible for propagation of wave through it ? |
| Answer» Correct Answer - Properties of elasticity and intertia | |
| 128. |
How does particle velocity differ from wave velocity? |
| Answer» Correct Answer - The particles veries both with position and time, whereras wave velocity for `a` wave motion remains the same. | |
| 129. |
What is the speed of trasverse waves in a wire, when stretched by a weight of `25 kg`? The two metre length of the wire has a mass of `4.7 g`? |
| Answer» Correct Answer - `322.9 ms^(-1)` | |
| 130. |
A steel wire 0.72 m long has a mass of `5.0 xx 10^(-3)` kg . If the wire is under a tension of 60 N, what is the speed of transverse waves on the wire ? |
| Answer» Correct Answer - `22.28 ms^(-1)` | |
| 131. |
Statement I: two waves moving in a uniform string having uniform tension cannot have different velocities. Elastic and inertial properties of string are same for all waves in same string. Moreover speed of wave in a string depends on its elastic and inertial properties only.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 Two waves moving in uniform string with uniform tension shall have same speed and may be moving in opposite direction. |
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| 132. |
A pulse is started at a time `t = 0` along the `+x` directions an a long, taut string. The shaot of the puise at `t = 0` is given by funcation `y` with `y = {{:((x)/(4)+1fo r-4ltxle0),(-x+1fo r0ltxlt1),("0 otherwise"):}` here `y` and `x` are in centimeters. The linear mass density of the string is `50 g//m` and it is under a tension of `5N`, The transverse velocity of the particle at `x = 13 cm` and `t = 0.015 s` will beA. `-250 cm//s`B. `-500 cm//s`C. `500 cm//s`D. `-1000 cm//s` |
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Answer» Correct Answer - A Transverse velocity `= (dely)/(delt)` at `t = 0.015 s, vt = 15 cm` as for `x = 13 cm(vt - 4) lt x lt vt` therfore `(dely)/(delt) = -(v)/(4) = -250 cm//s` |
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| 133. |
A pulse is started at a time `t = 0` along the `+x` directions an a long, taut string. The shaot of the puise at `t = 0` is given by funcation `y` with `y = {{:((x)/(4)+1fo r-4ltxle0),(-x+1fo r0ltxlt1),("0 otherwise"):}` here `y` and `x` are in centimeters. The linear mass density of the string is `50 g//m` and it is under a tension of `5N`, The vertical displacement of the particle of the string at `x = 7 cm` and `t = 0.01 s` will beA. `0.75 cm`B. `0.5 cm`C. `0.25 cm`D. zero |
| Answer» Correct Answer - C | |
| 134. |
Consider a sinusoidal travelling wave shown in figure. The wave velocity is `+ 40 cm//s`. Find (a) the frequency (b) the phase difference between points `2.5 cm` apart ( c ) how long it takes for the phase at a given position to (pi//3) (d) the velocity of a particle at `P` at the instant shown. A. `20 Hz`B. `30 Hz`C. `25 Hz`D. `10 Hz` |
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Answer» `lambda=2(4.5-2.5)=4 cm, v=nlambda` `rArr n=(40)/(4)=10 Hz` `phi=(2pi)/(lambda)Deltax=(5pi)/(4)rad implies phi=(2pi)/(T)Deltat` `Deltat=(phi)/(2pin)=(pi//3)/(2pi10)=(1)/(60)s` velocity of `p` should be maximum, as it is at mean position. `v_(p)=omegaA=2pifA=2pixx10xx(2)/(100)` `=1.26 m//s` this velocity should be `-ve`, because slope at `p` is `+ve`. |
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| 135. |
`y(x, t) = 0.8//[4x + 5t)^(2) + 5]` represents a moving pulse, where `x` and `y` are in meter and `t` in second. ThenA. pulse is moving in +x directionB. in `2` s it will travel a distance of `2.5 m`C. its maximum displacement is `0.16 m`D. it is a symmetric pulse |
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Answer» Correct Answer - b., c., d. `y(x,t)=(0.08)/(16[(x+(5)/(4)t)^(2)+(5)/(16)])` `y(x,t)=(0.05)/((x+(5)/(4)t)^(2)+(5)/(16))` where wave velocity = `5//a m//s` Distance travelled by the wave with this velocity in 2 s is `5//4(2)=2.5 m` `y(x,t)` will be maxium when `4x+5t=0` `y(max)=(0.8)/(5)=0.16 m` |
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| 136. |
The transverse displacement `y(x, t)` of a wave on a string is given by `y(x, t)= e ^(-(ax^(2) + bt^(2) + 2sqrt((ab))xt)`. This represents a :A. wave moving `+x-`direction with speed `sqrt((a)/(b))`B. wave moving in `-x`direction with speed `sqrt((b)/(a))`C. standing wave of frequency `sqrt(b)`D. standing wave of frequency `(1)/(sqrt(b))` |
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Answer» Correct Answer - 2 `y(x,t) = e^(-[sqrt(ax)+sqrt(bt)]^(2))` it is transverse type , `y(x,t) = e^(-(ax + bt)^(2))` Speed `v = (sqrt(b))/(sqrt(a))` and wave is moving along `-x` direction. `y(x,t) = e^(-[sqrt(ax)+sqrt(bt)]^(2))` |
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| 137. |
Distinguish between transverse and longitudinal wave motion. |
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Answer» Correct Answer - The wave motion is of two types namely transverse and logitudinal wave motiom 1. Transverse wave motion. When the particles of the medium vibrate about their mean position in a direction perpendicular to the direction of disturbance, the wave motion is called the transverse wave motion. 2. Longitudinal wave motion. When the particles of the medium vibrate about their mean position in the direction of propagation of disturbance, the wave motion is called the longitudinal wave motion. |
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| 138. |
which of the following function corrtly represent the traveling wave equation for finite values of x and t ?A. `y=x^(2)-t^(2)`B. `y=cosx^(2)sint`C. `y=log(x^(2)-t^(2))-log(x-t)`D. `y=e^(2x)sint` |
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Answer» Correct Answer - c |
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| 139. |
Which of the following function represent a travelling wave? Here a,b and c are constant.A. `y=a cos (bx) sin (ct)`B. `y=a sin (bx+ct)`C. `y=a sin (bx+ct)+a sin (bx-ct)`D. `y=a sin (bx-ct)` |
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Answer» Correct Answer - b., d. A travelling wave is characterized by wave functions of the type `y=f(vt+x) or y=f(vt-x)`. The function of the `sin (bx+ct)` represent a wave travelling in the negative x-direction and the function `y=a si (bx-ct)` a wave in positive x-direction. Hence, the correct choice are (b) and (d). |
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| 140. |
Given below are some functions of x and t to represent the displacement (transverse or longitudinal) of an elastic wave. State which of these represent (i) a travelling wave, (ii) a stationary wave or (iii) none at all? (a) `y=2cos(3x)sin(10t)` (b) `y=2sqrt(x-vt)` (c) `y=3sin(5x-0.5t)+4cos(5x-0.5t)` (d) `y=cosx sint +cos2x sin2t` |
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Answer» Correct Answer - (a) Stationary wave. (b) Unacceptable funcation for any wave. (c) Travelling harmonic wave. (d) Superposition of two stationary waves. |
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| 141. |
which of the following function corrtly represent the traveling wave equation for finite values of x and t ?A. `y = x^(2) - t^(2)`B. `y = cosx^(2) sint`C. `y = log(x^(2) - t^(2)) - log(x - t)`D. `y = e^(2x) sint` |
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Answer» Correct Answer - C Satisfy the standard equation of wave |
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| 142. |
A string is stretched betweeb fixed points separated by `75.0 cm`. It observed to have resonant frequencies of `420 Hz` and `315 Hz`. There are no other resonant frequencies between these two. The lowest resonant frequency for this strings isA. `10.5 Hz`B. `105 Hz`C. `1.05 Hz`D. `1050 Hz` |
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Answer» Correct Answer - 2 `(n + 1) (v)/(2l) = 420 …..(1)` `(nv)/(2l) = 315 ….(2)` `(1) - (2) (V)/(mu) = 105 Hz , f_(min) = = 105 Hz` |
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| 143. |
The equation of a travelling wave is `y60 cos(1800t-6x)` where y is in microns, t in seconds and x in metres. The ratio of maximum particle velocity to velocity of wave propagation isA. 3.6B. `3.6xx10^(-6)`C. `36xx10^(-11)`D. `3.6xx10^(-4)` |
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Answer» Correct Answer - d Maximum particle velocity =`omegaA` wave velocity =`(omega)/(K)` Therefore, the required ratio `=(omegaA)/(omega//K)` `AK` `=60xx10^(-6)xx6` `=3.6xx10^(-4)` |
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| 144. |
A heavy uniform rope is held vertically and is tensioned by clamping it to a rigid support at the lower end. A wave of a certain frequency is set up at the lower end. Will the wave travel up the rope with the same speed? |
| Answer» No. Due to the weight of the rope, the tension will increase along the string from the lower end to the upper end. Hence, the wave will travel with an increasing velocity along the string, since `v prop sqrtT` | |
| 145. |
The equation of a wave is `y=2simpi(0.5x-200t)` , where x and y are expressed in can and t in sec. The wave velocity is.......... |
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Answer» Correct Answer - `400cm//sec` |
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| 146. |
In a stationary wave represented by `y = a sin omegat cos kx`, amplitude of the component progerssive wave isA. `(a)/(2)`B. `a`C. `2a`D. None of these |
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Answer» Correct Answer - A `y = a sin omegat cos Kx` `y = (1)/(2) (2a sin omegat cos Kx)` `:.` Amplitude of component wave is `(a)/(2)` |
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| 147. |
A plane sound wave is travelling in a medium. In reference to a frame A, its equation is y=a cos `(omegat-kx)`. Which refrence to frame B, moving with a constant velocity v in the direction of propagation of the wave, equation of the wave will beA. `y =a cos[ omegat+kvt-kx]`B. `y =-a cos[omegat-kvt-kx]`C. `y=acos[omegat-kvt-kx]`D. `y =acos[ omegat+ kv t +kx]` |
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Answer» Correct Answer - c suppose at an instant t, the x-coordinate of a point with reference to moving frame is `x_(0)`. Since, at this moment, orgin of moving frame is at distance vt from origin of the fixed reference frame, therefore, putting this value of x in the given equation, we get `y=a cos [omegat-k(vt+x_(0))]` `y=a cos [omega-kv)t-x_(0)]` Hence, option (c ) is correct. |
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| 148. |
Figure below shows the wave `y=Asin (omegat-kx)` at any instant travelling in the `+ve` x-direction. What is the slope of the curve at B?A. `omega//A`B. k/AC. kAD. `omegaA` |
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Answer» Correct Answer - A |
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| 149. |
One end of a long string of linear mass density `10^(-2) kg m^(-1)` is connected to an electrically driven tuning fork of frequency 150 Hz. The other end passes over a pulley and is tied to a pan containig a mass of 90 kg. the pulley end absorbs all the incoming energy so that reflected waves from this end have neglidible amplitude , At t =0, the left end (fork end ) of the string is at x=0 has a transverse displacement of 2.5 cm and is moving along positive y-direction. the amplitude of the wave is 5 cm. write down the transverse displacement y (in cetimetres) as function of x(in metres) and t (in seconds ) that describes the wave on the string. |
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Answer» `theta=pi//6` `v=sqrt(T//mu)=sqrt(900)/(0.01)=300 m//s` frequency `=150 Hz` `:.lambda=(v)/(f)=(300)/(150)=2m` Then equation will be `y=A sin{2pi((t)/(T)-(x)/(lambda))+theta}` `=5 sin{2pi(150t-(x)/(2))+(pi)/(6)}` `=5 sin{pi(300t-x)+(pi)/(6)}` |
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| 150. |
A wave motion has the function `y=a_0sin(omegat-kx)`. The graph in figure shows how the displacement y at a fixed point varies with time t. Which one of the labelled points Shows a displacement equal to that at the position `x=(pi)/(2k)` at time `t=0`?A. PB. QC. RD. S |
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Answer» Correct Answer - b Given `y=a_(0) sin (omegat-kx)` Put t=0 and `x=(pi)/(2k)` `y=a_(0) sin (omegaxx0-kpi//2k)impliesy=-a_(0)` This corresponds to point `Q`. |
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