MCQ Questions for Class 11 Engineering Physics Waves Quiz 12

Three simple harmonic motions in the same direction having the same amplitude

$a$ and period are superposed. IF each differs in phase from the next by

${45}^{o}$ then

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the resultant amplitude $(1+\sqrt{2})a$
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the phase of the resultant motion relative to the first is ${90}^{o}$
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the energy associated with the resulting motion is $(3+2\sqrt{2})$ times the energy associated any single motion
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the resulting motion is not simple harmonic

Explanation

$\begin{array}{l} Let\, simple\, harmonic\, motions\, be\, represented\, by\, \\ { y_{ 1 } }=a\sin \left( { \omega t-\frac { \pi }{ 4 } } \right) \, \, ;\, \, { y_{ 2 } }=s\, \sin \omega t\, and \\ { y_{ 3 } }=a\sin \left( { \omega t+\frac { \pi }{ 4 } } \right) \\ On\, erimpo\sin g\, resul\tan t\, SHM\, will\, be \\ y=a\left[ { \sin \left( { \omega t-\frac { \pi }{ 4 } } \right) +\sin \omega t+\sin \left( { \omega t+\frac { \pi }{ 4 } } \right) } \right] \\ =a\left[ { 2\sin \omega t\cos \frac { \pi }{ 4 } +\sin \omega t } \right] \\ =a\left[ { \sqrt { 2 } \sin \omega t+\sin \omega t } \right] \\ =a\left( { 1+\sqrt { 2 } } \right) \sin \omega t \\ resul\tan t\, amplitude\, =\left( { 1+\sqrt { 2 } } \right) a \\ Now, \\ \frac { { { E_{ { { Re } }sul\tan t } } } }{ { { E_{ Single } } } } ={ \left( { \frac { A }{ a } } \right) ^{ 2 } }={ \left( { \sqrt { 2 } +1 } \right) ^{ 2 } }=\left( { 3+2\sqrt { 2 } } \right) \\ { E_{ { { Re } }sul\tan t } }=\left( { 3+2\sqrt { 2 } } \right) { E_{ S{ { ingle } } } } \end{array}$

Hence, the option

$C$ is the correct answer.

Equations of motion in the same direction are given by:

$y_1 = a \ sin \ (\omega t - kx)$

$y_2 = a \ sin \ (\omega t - kx - \theta)$
The amplitude of the medium particle will be:

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$2a\cos \frac{\theta }{2}$
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$\sqrt{2}a \ cos \ \theta$
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$4a \ cos \ \theta/2$
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$\sqrt{2}a \ cos \ \theta/2$

Explanation

$\begin{array}{l} The\, \, resul\tan t\, \, amplitude\, \, is\, \, given\, \, by \\ { a_{ R } }=\sqrt { { a^{ 2 } }+{ a^{ 2 } }+2\cdot a\cdot a\cos \theta } \\ =\sqrt { 2{ a^{ 2 } }\left( { 1+\cos \theta } \right) } \\ =2a\cos \frac { \theta }{ 2 } \, \, \, \, \, \, \, \, \, \, \, \, \left( { H\cos \theta =2{ { \cos }^{ 2 } }\frac { \theta }{ 2 } } \right) \end{array}$

Hence,

option

$(A)$ is correct answer.

Equations of motion in the same direction are given by:

$y_1 = 2a \ sin \ (\omega t - kx)$

$y_2 = 2a \ sin \ (\omega t - kx - \theta)$
The amplitude of the medium particle will be:

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$2a\cos \frac{\theta }{2}$
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$\sqrt{2}a \ cos \theta$
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$4a \ cos \theta/2$
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$\sqrt{2}a \ cos \theta/2$

Explanation

Hence,

option

$(A)$ is correct answer.

A small mass executes linear SHM about

$O$ with amplitude

$a$ and period

$T$ . Its displacement from

$O$ at time

$T/S$ after passing through

$O$ is

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$a/S$
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$\cfrac{a}{2\sqrt{2}}$
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$a/2$
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$\cfrac{a}{\sqrt{2}}$

Explanation

$\begin{array}{l} y=a\sin wt \\ =a\sin \frac { { 2\pi } }{ T } .\frac { T }{ 4 } \\ =a\sin \frac { \pi }{ 4 } =\frac { a }{ { \sqrt { 2 } } } \end{array}$

Hence, the option

$D$ is the correct answer.

If the fundamental frequency of string is

$220 \mathrm { cps }$ , the frequency of fifth harmonic will be

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$44\mathrm { cps }$
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$55 \mathrm { cps }$
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$1100 \mathrm { cps }$
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$440 \mathrm { cps }$

A particle performs simple harmonic motion with amplitude A. Its speed is trebled at the instant that it is at a distance

$\frac{2A}{3}$ from equilibrium position. The new amplitude of the motion is

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$\dfrac{A}{3}\sqrt {41}$
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3A
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$A\sqrt3$
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$\dfrac{7A}{3}$

Explanation

$\begin{array}{l} m{ w^{ 2 } }=k \\ Total\, \, initial\, \, energy=\frac { 1 }{ 2 } k{ A^{ 2 } } \\ at\, x=\frac { { 2A } }{ 3 } ,\, \, potential\, \, energy=\frac { 1 }{ 2 } k{ \left( { \frac { { 2A } }{ 3 } } \right) ^{ 2 } } \\ =\left( { \frac { 1 }{ 2 } k{ A^{ 2 } } } \right) \left( { \frac { 4 }{ 9 } } \right) \\ Kinetic\, \, energy\, \, at\left( { x=\frac { { 2A } }{ 3 } } \right) \, \, is=\left( { \frac { 1 }{ 2 } k{ A^{ 2 } } } \right) \left( { \frac { 5 }{ 9 } } \right) \\ If\, \, speed\, \, is\, \, triled,\, \, new\; \; kinetic\, \, energy=\frac { 1 }{ 2 } k{ A^{ 2 } }\cdot \frac { 5 }{ 9 } =\frac { 5 }{ 2 } k{ A^{ 2 } } \\ \therefore New\, \, total\, \, energy\, \, =\frac { 5 }{ 2 } k{ A^{ 2 } }+\frac { 1 }{ 2 } k{ A^{ 2 } }\left( { \frac { 4 }{ 9 } } \right) \\ if\, \, next\, \, amplitude\, \, =A' \\ then\, \, \frac { 1 }{ 2 } kA{ '^{ 2 } }\left( { \frac { { 49 } }{ 9 } } \right) \\ \Rightarrow A'=\frac { { 7A } }{ 3 } \end{array}$

Hence,

option

$(D)$ is correct answer.

A particle executes a SHM of time T, find the time taken by the particle to go directly from its mean position to half the amplitude

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$\dfrac{T}{2}$
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$\dfrac{T}{8}$
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$\dfrac{T}{12}$
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$\dfrac{T}{4}$

The phase (at a time t) of a particle in simple harmonic motion tells

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Only the position of the particle at time t
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Only the direction of motion of the particle at time t
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Both the position and direction of motion of the particle at time t
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Neither the position of the particle nor its direction of motion at time t

The graph plotted between the velocity and displacement from mean position of a particle executing S,H.M. is

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circle
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ellipse
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parabola
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straight line.

The amplitude of a wave represented by displacement equation

$y=\frac{1}{\sqrt{a}}$

$\sin \omega t$

$\pm$

$\frac{1}{\sqrt{b}}$

$\cos \omega t$ will be

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$\frac{a+b}{a b}$
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$\frac{\sqrt{a}+\sqrt{b}}{a b}$
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$\frac{\sqrt{a} \pm \sqrt{b}}{a b}$
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$\sqrt{\frac{a+b}{a b}}$

Explanation

The correct option is D

Hint: Using trigonometry, convert the given function in suitable form.

Explanation for correct answer:

Step 1: Information required,

The displacement equation is given as,

$y = \dfrac{1}{{\sqrt a }}\sin \omega t \pm \dfrac{1}{{\sqrt b }}\cos \omega t$

Reworking the wave condition as $\sin$ . Change the $\cos$ to $\sin$ by $\cos \omega t$ = $\sin \left( {\omega t + \dfrac{\pi}{2}} \right)$

$y = \dfrac{1}{{\sqrt a }}\sin \omega t \pm \dfrac{1}{{\sqrt b }}\sin \left( {\omega t + \dfrac{\pi }{2}} \right)………….(1)$

From the above equation it is observed that the phase difference is $\dfrac{\pi }{2}$ .

Step 2: Consider general wave equation,

The general wave equation is given as,

$y = A\sin \left( {\omega t + \theta } \right)…………….(2)$

Where A represents the amplitude.

Compare amplitudes in both the equations (1) and (2),

Consider amplitudes as ${A_{1,}}{A_2}$

Then,

${A_1} = \dfrac{1}{{\sqrt a }}$ , ${A_2} = \dfrac{1}{{\sqrt b }}$

Resultant amplitude is given as,

$A = \sqrt {{A_1}^2 + {A_2}^2}$

Substituting the ${A_{1,}}{A_2}$ values in the above equation, we get,

$A = \sqrt {{{(\frac{1}{{\sqrt a }})}^2} + {{(\frac{1}{{\sqrt b }})}^2}}$

$\Rightarrow A = \sqrt {\dfrac{1}{a} + \dfrac{1}{b}}$

$\therefore A = \sqrt {\dfrac{{a + b}}{{ab}}} \\$

Therefore, the amplitude of the wave is $A = \sqrt {\dfrac{{a + b}}{{ab}}} \\$ .

A particle performs SHM of amplitude

$A$ along a straight line. When it is at a distance

$\cfrac{\sqrt{3}}{2}A$ from mean position, its kinetic energy gets increased by an amount

$\cfrac{1}{2}m \omega {A}^{2}$ due to an impulsive force. Then its new amplitude becomes.

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$\cfrac{\sqrt{5}}{2}A$
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$\cfrac{\sqrt{3}}{2}A$
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$\sqrt{2}A$
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$\sqrt{5}A$

Explanation

$\begin{array}{l} If\, \, the\, particle\, \, performs\, \, SHM\, \, with\, \, amplitude\, \, A\, \, along\, \, the\, \, straight\, \, line,\, \, than\, \, at\, \, any\, \, po{ { int } }\, \, the\, \, total\, \, energy\, \, should\, \, be, \\ E=K+P=\frac { 1 }{ 2 } K{ A^{ 2 } }=\frac { 1 }{ 2 } m{ w^{ 2 } }{ A^{ 2 } } \\ If\, \, the\, additional\, \, kinetic\, \, energy\, \, of\, \, magnitude\, \, \frac { 1 }{ 2 } m{ w^{ 2 } }{ A^{ 2 } }\, \, by\, \, means\, \, of\, \, an\, \, impulsive\, \, force, \\ then\, \, the\, \, total\, \, energy\, \, of\, \, the\, \, particle\, \, changes\, \, to, \\ E=E+\frac { 1 }{ 2 } m{ w^{ 2 } }{ A^{ 2 } }=\frac { 1 }{ 2 } m{ w^{ 2 } }{ A^{ 2 } }+\frac { 1 }{ 2 } m{ w^{ 2 } }{ A^{ 2 } } \\ \frac { 1 }{ 2 } m{ w^{ 2 } }A{ '^{ 2 } }=\frac { 1 }{ 2 } m{ w^{ 2 } }\left( { 2{ A^{ 2 } } } \right) \\ \Rightarrow A{ '^{ 2 } }=2{ A^{ 2 } } \\ \Rightarrow A'=\sqrt { 2 } A \end{array}$

Hence,

option

$(C)$ is correct answer.

The amplitude of vibration of a particle is given by

${ a }_{ m }=\dfrac { { a }_{ 0 } }{ { a }w^{ 2 }-bw+c }$ Where

${ a }_{ 0 },a,b$ and

$c$ are positive. The condition for a single resultant frequency is

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${ b }^{ 2 }=4ac$
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${ b }^{ 2 }>4ac$
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$\\ { b }^{ 2 }=5ac\\$
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$\\ { b }^{ 2 }=7ac$

Explanation

For resonance, amplitude must be maximum which is possible only when the denominator of expression is zero i.e.

$a \omega^2-b \omega +c=0$

$\Rightarrow \omega$

$=\dfrac{+b \pm \sqrt{b^2-4ac}}{2a}$
For a single resonant frequency,

$b=4ac$

A particle performs S.H.M. of amplitude A along a straight line. When it is at a distance

$\frac{\sqrt 3}{2}$ A form measure position, its kinetic energy gets increased by an amount

$\frac{1}{2}$ m

$\omega^2A^2$ due to an impulsive force. Then new amplitude becomes:

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$\frac{\sqrt 5}{2}A$
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$\frac{\sqrt 3}{2}A$
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$\sqrt2 A$
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$\sqrt5 A$

Explanation

$\begin{array}{l} Given, \\ Dis\tan ce\, \, x=\frac { { \sqrt { 3 } } }{ 2 } \\ Increased\, kinetic\, energy\, \, K.E=\frac { 1 }{ 2 } m{ \omega ^{ 2 } }{ a^{ 2 } } \\ now, \\ E=K.E+P.E \\ E=\frac { 1 }{ 2 } m{ \omega ^{ 2 } }{ x^{ 2 } }+\frac { 1 }{ 2 } m{ \omega ^{ 2 } }\left( { { a^{ 2 } }-{ x^{ 2 } } } \right) \\ E=\frac { 1 }{ 2 } m{ \omega ^{ 2 } }{ a^{ 2 } } \\ When\, particle\, is\, at\, a\, dis\tan ce\, of\, \frac { { \sqrt { 3 } } }{ 2 } a\, from\, mean\, position,\, its\, kinetic\, energy\, gets\, \\ increased\, by\, an\, amount\, \frac { 1 }{ 2 } m{ \omega ^{ 2 } }{ a^{ 2 } }\, due\, to\, an\, impulsive\, force. \\ So,\, the\, new\, total\, energy\, is\, \\ E'=E+\frac { 1 }{ 2 } m{ \omega ^{ 2 } }{ a^{ 2 } } \\ \frac { 1 }{ 2 } m{ \omega ^{ 2 } }a{ '^{ 2 } }=\frac { 1 }{ 2 } m{ \omega ^{ 2 } }{ a^{ 2 } }+\frac { 1 }{ 2 } m{ \omega ^{ 2 } }{ a^{ 2 } } \\ a'=\sqrt { 2 } a \\ Hence,\; the\, option\, C\, is\, the\, correct\, answer. \end{array}$

Two equations of two SHM are

$x=a \ sin(\omega t - \alpha) \ and \ y=b \ cos(\omega t - \alpha)$
The phase difference between the two is

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$0^o$
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$\alpha^o$
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$90^o$
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$180^o$

Explanation

$\begin{array}{l} Equation\; \, for\, \, \, volatage\, \, is\; \, given\, as: \\ V={ V_{ 0 } }\cos \omega t \\ Equation\, \, for\, \, \, current\, \, is\, \, given\, as\, \, : \\ l={ l_{ .0 } }\cos \left( { \omega t-\phi } \right) \\ Where, \\ \phi =phase\, \, difference\, \, between\, \, voltage\, \, \, and\, \, \, currect \\ At\, \, time,t=0 \\ V={ V_{ 0 } }\left( { voltage\, \, is\, \, \, \max imum } \right) \\ For\, \, \omega t-\phi =0 \end{array}$

Two particles are executing SHM of same amplitude and frequency along the same straight line path. They pass each other when going in opposite directions, each time their displacement is half of their magnitude. What is the phase difference between them?

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$\pi / 6$
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$4 \pi / 6$
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$\pi / 3$
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$2 \pi / 6$

Explanation

$\begin{array}{l} We\, have \\ x=A\sin \left( { \omega t+\delta } \right) ;\, \, where\, \delta \, is\, the\, phase\, cons\tan t. \\ For\, particle\, 1 \\ at\, X=A\, and\, t=0 \\ \frac { A }{ 2 } =A\sin \delta \\ \Rightarrow \frac { 1 }{ 2 } =\sin \delta \\ \Rightarrow \delta =\frac { \pi }{ 6 } ,\frac { { 5\pi } }{ 6 } \\ The\, velocity\, is \\ v=\frac { { dx } }{ { dt } } =A\, \omega \cos \left( { \omega t+\delta } \right) \\ At\, t=0\, ,\, v=A\omega t\cos \delta ]Now, \\ \cos \left( { \frac { \pi }{ 6 } } \right) =\frac { { \sqrt { 3 } } }{ 2 } \, and\, \cos \left( { \frac { { 5\pi } }{ 6 } } \right) =-\frac { { \sqrt { 3 } } }{ 2 } \\ As\, v\, is\, positive\, for\, particle\, 1,\delta \, must\, be\, equal\, to\, \frac { \pi }{ 6 } \\ for,\, particle\, 2,\, V\, is\, negative\, ,\, \delta \, must\, be\, equal\, to\, \frac { { 5\pi } }{ 6 } \\ Thus,\, the\, phase\, difference\, between\, them\, is\, =\frac { { 5\pi } }{ 6 } -\frac { \pi }{ 6 } \\ =\frac { { 4\pi } }{ 6 } \\ Hence,\, the\, option\, B\, is\, the\, correct\, answer. \end{array}$

A particle moving along a straight line with a constant acceleration of - 4 m per second square passes through a point on the line with a velocity of + 8 metre per second at the moment when the distance travelled by the particle in 6 second in 5 seconds

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$5 m/s^2$
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$6 m/s^2$
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$8 m/s^2$
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$2 m/s^2$

Explanation

$uy=10sin 60=53√m/s$

$⇒t=2uyg=2×53√10=3√s$

$Sx=uxt+12axt^2$

$1.15=5×t−12a×t^2$

$1.15=5×3√−32a$

$3a^2=5×1.73−1.15=8.65−1.15$

$3a^2=7.5$

$⇒a=153=5m/s^2$

Four waves are expressed as
(i)

$y _ { 1 } = a _ { 1 } \sin \omega t$ (ii)

$y _ { 2 } = a _ { 2 } \sin 2 \omega t$
(iii)

$y _ { 3 } = a _ { 3 } \cos \omega t$ (iv)

$y _ { 4 } = a _ { 4 } \sin ( \omega t + \phi )$
The interference is possible between

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(i) and (iii)
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(i) and (ii)
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(ii) and (iv)
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Not possible at all

A wave equation which given the displacement along the Y direction is given by

$y = 10^{-4} \sin (60t+2x)$ where x and y are in meters and t is time in seconds. This represents a wave

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Traveling with a velocity of 30 m/s in the negative x direction
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Of wavelength $\pi$ metre
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Of frequency 30/ $\pi$ hertz
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Of amplitude $10^{ -4 }$ metre

Displacement of a particle performing S.H.M. is given by

$x = 0.01$

$\sin \pi ( t + 0.05 ) ,$ where

$x$ is in meter and t is in seconds. The time period in second is :

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$2$
0%
$0.01$
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$0.02$
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$0.1$

Explanation

Hence, option

$(C)$ is correct answer.
1380926_1535132_ans_00afb3c591b54d6bb4d79b98801aa8d4.jpg

The displacement equation of a simple harmonic oscillator is given by

$x = 6 \sin \left[\dfrac{\pi}{8} + \dfrac{\pi}{16}t\right] m$ when t = 2 seconds is

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1.3m
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1.5m
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$3\sqrt{3} m$
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$3 \sqrt{2}$ m

On the superposition of the two waves given as

${y}_{1}={A}_{0}\sin{(\omega t-kx)}$ and

${y}_{2}={A}_{0}\cos {(\omega t-kx+\cfrac{\pi}{6}})$ the resultant amplitude of oscillation will be

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$\sqrt{3}{A}_{0}$
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$\cfrac{{A}_{0}}{2}$
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${A}_{0}$
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$\cfrac{3}{2}{A}_{0}$

Explanation

Given,

$y_1=A_0 sin(\omega t-kx)$

$y_2=A_0 cox(\omega t-kx+\dfrac{\pi}{6})=A_0 sin(\omega t-kx+2\pi /3)$

$A_1=A_0$

$A_2=A_0$

$\theta=2\pi/3$

Resultant amplitude,

$R=\sqrt{A_1^2+A_2^2+2A_1A_2cos\theta}$

$R=\sqrt{A_0^2+A_0^2+2A.Acos120^0}$

$R=\sqrt{2A_0^2-A^2}$

$R=A_0$

The resultant amplitude is

$A_0$ .

The correct option is C.

The displacement of a particle ( in meters ) from its mean position is given by the equation $y = 0.2\left( {{{\cos }^2}\dfrac{{\pi t}}{2} - {{\sin }^2}\dfrac{{\pi t}}{2}} \right)$ . The motion of the above particle is.

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not simple harmonic
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Simple harmonic with amplitude of 0.2 m
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Simple harmonic with a period equal double that of a second's pendulum
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Simple harmonic with an amplitude of 0.4 m

Figure given showsa sinusoidal wave on a string.If the frequency of the wave is 150 Hz and the mass per unit length of the string is 0.2 g/s.the power transmitted by the wave is

0%
2.34 W
0%
3.84 W
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4.80 W
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5.78 W

Equation

${ y }_{ 1 }=0.1sin\left( 100\pi t+\dfrac { \pi }{ 3 } \right)$ and

${ y }_{ 2 }=0.1$ cos

$\pi t$ The phase difference of the velocity of particle 1, with respect to the velocity of particle 2 is

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$\dfrac { -\pi }{ 6 }$
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$\dfrac { \pi }{ 3 }$
0%
$\dfrac { -\pi }{ 3 }$
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$\dfrac { \pi }{ 6 }$

Number of waves of a light of wavelength

$5000 \mathrm { A } ^ { \circ } ,$ in a thickness of

$0.2 \mathrm { mm } ,$ of air is

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$100$
0%
$200$
0%
$250$
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$400$

The equation of a progressive wave is

$Y= a sin(200 t-x)$ , where x is in meter and t is in second. The velocity of wave is

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$200$ m/sec
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$100$ m/sec
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$50$ m/sec
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None

The equation of a wave is given by

$Y\quad =\quad A\quad sin\quad \omega \left( \frac { x }{ v } -k \right)$
Where

$\omega$ is the angular velocity and v is the linear velocity.The dimensions of K is

0%
LT
0%
T
0%
$T^{-1}$
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$T^2$

Explanation

From the principle of homogenity

$(\dfrac{x}{v})$ has dimensions of T.

Two points are located at a distance of

$10\ m$ and

$15\ m$ from the source of oscillation. The period of oscillation is

$0.05\sec$ and the velocity of the wave is

$300\ m/\sec$ . What is the phase difference between the oscillations of two points?

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$\dfrac {\pi}{6}$
0%
$\dfrac {\pi}{3}$
0%
$\dfrac {2\pi}{3}$
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$\pi$

Which of the following figure about reflection of transverse or longitudinal waves is CORRECT?

$(V_w = velocity \ of \ wave, v_p = particle\ velocity)$

The equation

$y =A\cos^2\left(2\pi\, nt -2\pi \dfrac{x}{\lambda}\right)$ represents a wave with

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amplitude $A/2$ , frequency $2n$ & wavelength $\lambda/2$
0%
amplitude $A/2$ , frequency $2n$ & wavelength $\lambda$
0%
amplitude $A$ , frequency $2n$ & wavelength $2\lambda$
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amplitude $A$ , frequency $n$ & wavelength $\lambda$

For a wave

$y= y_0 sin ( \omega t - kx )$ , for what value of

$\lambda$ , is the maximum particle velocity equal to two times the wave velocity :-

0%
$\pi y_0$
0%
$2 \pi y_0$
0%
$\pi y_0/2$
0%
$4 \pi y_0$

Which of the following statement is incorrect superposition of waves?
(i) After superposition frequency,wavelength and velocity of resultant wave remains same
(ii) After superposition amplitude of resultant wave is equal to amplitude of either wave
(iii) Mechanical wave cannot superposed with electromagnetic wave
(iv) For superposition two waves should have equal wavelength,frequency and amplitude

0%
I & III Only
0%
I & II Only
0%
I,II, & III Only
0%
II & IV Only

Explanation

If both the waves have same frequency wavelength and velocity then only superposition frequency, wavelength and velocity of resultant wave remains same.

Superposition amplitude of resultant wave is given by,

$A = \sqrt{A_1^2+A_2^2 +2A_1A_2\cos{\theta}}$ , which may of may not be equal to amplitude of either wave.

The equation

$y = a \sin^2 \left(2 \pi nt - \dfrac{2\pi x}{\lambda}\right)$ represents a wave with

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Amplitude $a$ , frequency $n$ and wavelength $\lambda$
0%
Amplitude $a$ , frequency $2n$ and wavelength $2\lambda$
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Amplitude $a/2$ , frequency $2n$ and wavelength $\lambda$
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Amplitude $a/2$ , frequency $2n$ and wavelength $\lambda/2$

The displacement equation of a progressive wave is y = 3 sin (200 t-5 x), Where the distance x and y are in meter and time t in second. The speed of the wave is:

0%
$20 m s^{-1}$
0%
$40 m s^{-1}$
0%
$60 m s^{-1}$
0%
$80 m s^{-1}$

Explanation

Given:

$y = 3\sin(200t-5x)$

Comparing it with standard wave equation,

$y = A \sin(\omega t -kx)$

$\omega = 200$

$s^{-1}$

$k = 5$

$m^{-1}$

Velocity of wave is given by,

$v =\dfrac{\omega}{k}$

$v = \dfrac{200}{5} = 40$

$m/s$

If the amplitude of a wave at a distance

$r$ from a point source is

$A$ , the amplitude at a distance

$2r$ will be

0%
$2A$
0%
$A$
0%
$A/2$
0%
$A/4$

Explanation

Intensity,

$I \propto (A)^2$

Also,

$I \propto \dfrac{1}{r^2}$

So,

$A \propto \dfrac{1}{r}$

At distance r, amplitude is A, so at distance 2r amplitude will be

$\dfrac{A}{2}$

The phase difference between two particles executing SHM of the same amplitudes and frequency along same straight line while passing one another when going in opposite directions with equal displacement from their respective starting point is

$2\pi/3$ . If the phase of one particle is

$\pi/6$ , find the displacement at this instant, if amplitude is

$A$ .

0%
$A/3$
0%
$2A/3$
0%
$3A/4$
0%
$A/2$

At any instant a wave travelling along the string is shown in fig. Here, point

$'A'$ is moving upward. Which of the following statements is true?

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The wave is travelling to the right
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The displacement amplitude of wave is equal to displacement of $B$ at this instant
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At this instant $'C'$ also directed upward
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None of these

A wave has a wavelength of 3m. The distance between a crest and adjacent trough is

0%
0.75 m
0%
1.5 m
0%
3 m
0%
1 m

Explanation

Distance between two adjacent crest or trough is wavelength, and distance between a crest and adjacent trough is half of wavelength.

Given,

$\lambda = 3m$

Distance between a crest and adjacent trough

$= \dfrac{\lambda}{2} = \dfrac32 = 1.5m$

A wave pulse in a string is described by the equation

$y_1=\dfrac{5}{(3x-4t)^2+2}$ and another wave pulse in the same string is described by

$y_2=\dfrac{-5}{(3x+4t-6)^2+2}$ . The values of

$y_1,y_2$ and

$x$ are in metres and

$t$ in seconds. Which of the following statement is correct?

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$y_1$ travels along -x- direction and $y_2$ along +x-direction
0%
Both $y_1$ and $y_2$ travel along -x-direction
0%
Both $y_1$ and $y_2$ travel along +x-direction
0%
At $x=1m,y_1$ and $y_2$ always cancel
0%
At time $t=1s,y_1$ and $y_2$ exactly cancel everywhere

Explanation

$y_1=\dfrac{5}{(3x-4t)^2+2}, y_2=\dfrac{-5}{(3x+4t-6)^2+2}$

According option (d), at

$x=1$

$y_1=\dfrac{5}{(3-4t)^2+2}$

$y_2=\dfrac{-5}{(3+4t-6)^2+2}$

$=\dfrac{-5}{(3-4t)^2+2}$

Both wave pulse equation are existing in same string therefore resultant equation of wave pulse.

$y=y_1+y_2=0$

$x=1m,y_1$ and

$y_2$ always cancel.
So, the correct option is

$(D)$

A wave pulse moves along a stretched rope in the direction shown.
Which diagram shows the variation with time t of the displacement s of the particle P in the rope?

The wavelength of the first line of Lyman series is

$\lambda$ . The wavelength of the first line in Paschen series is ________.

0%
$108/7$
0%
$27/5$
0%
$7/108$
0%
$5/27$

Explanation

Formula for Lyman series is:

Where, n = 2,3,4,5,.....

Where, R = Rydbergg constant,

$\dfrac{1}{\lambda}=R\left(\dfrac{1}{1^2}-\dfrac{1}{2^2}\right)$ and

Paschen series in

$\lambda_1$ ,

$\Rightarrow \dfrac{1}{\lambda_1}=R\left(\dfrac{1}{3^2}-\dfrac{1}{4^2}\right)$

$\Rightarrow \dfrac{\lambda_1}{\lambda}=\dfrac{\dfrac{3}{4}}{\dfrac{7}{16\times 9}}$

$\Rightarrow \lambda_1=\dfrac{3}{4}\times \dfrac{16\times 9}{7}\lambda$

$\Rightarrow \lambda_1=\dfrac{108}{7}\lambda$ .

A sinusoidal wave travelling in the same direction have amplitude of

$3\ cm$ and

$4\ cm$ and difference in phase by

$\pi/2$ . The resultant amplitude of the superimposed wave is :

0%
$7\ cm$
0%
$5\ cm$
0%
$2\ cm$
0%
$0.5\ cm$

The equation of standing wave is

$y=a\cos kx \sin \omega t$ which one of following graphs is for the wave at

$t=\dfrac{T}{4}$ ?

0%
0%
0%
0%
none of the above

The number of waves each of wavelength

$10\ cm$ produced in string of

$100\ cm$ length, is:

0%
$1$
0%
$10$
0%
$20$
0%
$30$

Equation of a plane wave is given by

$4\sin \dfrac{\pi}{4}\left[2t+\dfrac{x}{8}\right]$ . The phase difference at any given instant of two particles

$16\ cm$ apart is :

0%
$60^o$
0%
$90^o$
0%
$30^o$
0%
$120^o$

Wave of frequency

$500\ Hz$ has a phase velocity

$360\ m/s$ . The phase difference between two displacement at a certain point at time

$10^{-3}\ s$ apart will be :

0%
$\pi$ radian
0%
$\dfrac{\pi}{2}$ radian
0%
$\dfrac{\pi}{4}$ radian
0%
$2\ pi$ radian

Velocity of sound waves in air is

$330\ m/s$ . For a particular sound in air, a path difference of

$40\ cm$ is equivalent to a phase difference of

$1.6\ \pi$ . The frequency of this wave is:

0%
$165\ Hz$
0%
$150\ Hz$
0%
$660\ Hz$
0%
$330\ Hz$

The phase change between incident and reflected sound wave from a fixed wall is:

0%
$0$
0%
$\pi$
0%
$3\pi$
0%
$\dfrac{\pi}{2}$

Two waves represented by

$y = a \sin (\omega t - kx)$ and

$y = a \cos (\omega t - kx)$ are superposed. The resultant waves will have an amplitude.

0%
$a$
0%
$\sqrt{2a}$
0%
$2a$
0%
$0$

Two point lie on a ray are emerging from a source of simple harmonic wave having period

$0.045$ . The wave speed is

$300\ m/s$ and points are at

$10\ m$ and

$16\ m$ from the source. They differ in phase by :

0%
$\pi$
0%
$\pi/2$
0%
$0$ or $2\pi$
0%
$none\ of\ these$

CBSE Questions for Class 11 Engineering Physics Waves Quiz 12 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Waves Quiz 12 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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