MCQ Questions for Class 11 Engineering Physics Waves Quiz 9

The refracted and the incident pulses for a wave travelling in a string have an amplitude ratio of 1:The ratio of their phases will be

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1:2
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2:1
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1:1
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1:4

Principle of superposition can be applied to

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EM waves
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sound waves
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Radio waves
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All of the above

A progressive wave

$y = 10 sin (2t-4x)$ gets reflected along a rigid boundary as

$y= 8 sin (2t+4x+\pi)$ under certain conditions

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True
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False

Two waves of same amplitude and frequency and travelling in same direction are superimposed each other to give rise to a resultant wave of amplitude 2A

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True
0%
False

Explanation

The resultant of two waves is given by

$R^2=A^2+B^2+2AB cos \theta; \theta$ is the phase difference and A and B are the amplitudes of the individual waves.

Since both the waves are identical, their amplitudes are equal

Thus, R = A+A=2A

If the phase difference between two component waves of different amplitudes is

$2\pi$ , their resultant amplitude will become

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sum of the amplitudes
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difference of the amplitudes
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product of the amplitudes
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ratio of their amplitudes

Explanation

The resultant of two waves is given by

$R^2=A^2+B^2+2AB cos \theta; \theta$ is the phase difference and A and B are the amplitudes of the individual waves.

If

$\theta = 2\pi$ , then

$R = A+B$

The correct option is (a)

The initial phase of a pulse travelling on a string is

$\pi/3$ . The reflected pulse has a phase of

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$4 \pi/3$
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$2 \pi/3$
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$\pi/3$
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$5 \pi/3$

Explanation

The incident and reflected pulse are

$\pi$ radians out of phase. Thus reflected ray will have a phase of

$\pi + \pi/3=4 \pi/3$

The correct option is (a)

The phase of the transmitted wave changes with frequency

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True
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False

The equation of an incident wave travelling along +X direction is given by

$y= A sin (2t-5x)$ . This wave gets reflected at a rigid boundary. The equation of the reflected wave is

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$y= A sin (2t-5x)$
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$y= A sin (2t-5x +\pi)$
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$y= A sin (2t+5x +\pi)$
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$y= A sin (2t-5x +\pi/2)$

Explanation

A reflected wave will be

$\pi$ radians out of phase with the incident wave. Apart from this, the wave will also get reflected along the negative x axis. Thus, the sign of x also changes.

The equation of the reflected wave is thus

$y= A sin (2t+5x +\pi)$
The correct option is (c)

The principle of superposition may be applied to waves whenever two (or more) waves travelling through different media have same frequency

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True
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False

Two waves represented by

$y_1=10\sin(2000\pi t)$ and

$y_2=10\sin (2000 \pi t+\pi /2)$ are superimposed at any point at a particular instant. The resultant amplitude is?

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$10$ units
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$20$ units
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$14.1$ units
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Zero

Explanation

The resultant amplitude A of two waves of amplitudes

$a_1$ and

$a_2$ at a phase difference of

$\phi$ is

$(a^2_1+a^2_2+2a_1a_2\cos\phi)^{1/2}$ . Substituting

$a_1=10, a_2=10$ and

$\phi =90^o$ ,
we get

$A=14.1$
Hence (C) is correct.

Reflection at a free boundary implies

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restoring force is zero
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restoring force is not zero
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applied force is zero
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applied force is not zero

A wave travelling on a string as shown in figure gets reflected at an open boundary. Then,

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Phase reversal takes place at the boundary
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Wavelength changes at the boundary
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Phase reversal dosent take place
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Speed of the reflected ray takes place

The angular frequency of a particle in a progressive wave in an elastic medium is

$100\pi\ { rads }^{ -1 }$ and it is moving with a velocity of

$200{ms}^{-1}$ . The phase difference between two particles separated by a distance of

$20m$ is:

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$31.4\ rad$
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$\pi\ rad$
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$\dfrac{3\pi}{4}\ rad$
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$36\ rad$

The equation of a progressive wave is given by

$y=10 sin (5t-x)$ . The wave gets reflected from a open boundary. The equation of the reflected wave is

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$y=10 sin (5t-x)$
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$y=-10 sin (5t-x)$
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$y=10 sin (5t+x)$
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$y=10 sin (5t+x+\pi)$

${y}_{1}=5(mm)]sin{\pi t}$ and that of

${S}_{2}$ is

${y}_{2}=5(mm)\sin{(\pi t+\pi /6)}$ . A wave from

${S}_{1}$ reaches point A in

$1\ sec$ while a wave from

${S}_{2}$ reaches point A in

$0.5s$ . The resulting amplitude at point

$A$ is :-

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$5\sqrt { 2+\sqrt { 3 } } mm$
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$5\sqrt { 3 } mm$
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$5mm$
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$5\sqrt { 2 }mm$

Explanation

$Y_{1}=5 (mm) \sin \pi t$

$Y_{2}=5 (mm)\sin \left(\pi t+\dfrac{\pi}{6}\right)$

from both transverse wave equation

we can get

$A_{1}=5\ mm$

$A_{2}=5\ mm$

$\phi=(\pi t)+\pi/6-\pi t)=\dfrac{\pi}{6}$

where,

$A_{1}$ and

$A_{2}$ are amplitude of wave equation and

$\phi$ is the phase difference

resultant amplitude

$=\sqrt{A_{1}^{2}+A_{2}^{2}+2A_{1}A_{2}\cos\phi}$

$=5\sqrt{1+1+2\times 1\times 1\times \dfrac{\sqrt{3}}{2}}$

$=5\sqrt{2+\sqrt{3}}$

Resultant Amplitude

$=5\sqrt{2+\sqrt{3}}mm$

Correct option is

$(A)$

Displacement time equation of a particle executing SHM is x=4 sin

$\omega$ t + 3 sin (

$\omega t+ \dfrac{\pi}{3})$ here x is in centimeter and t is in seconds . The amplitude of oscillation of the particle is approximately:

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5 cm
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6 cm
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7 cm
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9 cm

Explanation

Resultant amplitude of given expression is A.

$A=\sqrt{A_1^2+A_2^2+2A_1A_2\cos\phi}$

$=\sqrt{4^2+3^2+2\times 4\times 3\cos \dfrac{\pi}{3}}$

$=\sqrt{16+9+12}=\sqrt{37} \cong6cm$

Two waves of frequencies 20 Hz and 30 Hz travels out from a common point. The phase difference between them after 0.6 sec is

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$12\pi$
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${\pi \over 2}$
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$\pi$
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${{3\pi } \over 4}$

Explanation

Beats period

$=\cfrac{1}{30-20}=0.1second\\ \Delta Y=\cfrac{2\pi}{T}\Delta t\\ \quad=\cfrac{2\pi}{0.1}\times0.6=2\pi\times6=12\pi\\ \therefore\Delta Y=12\pi$

The displacements of two particles executing

$SHM$ on the same line are given as

$y_{1}=a\sin \left (\dfrac {\pi}{2}t+ \phi \right)$ and

$y_{2}=b\sin \left (\dfrac {2\pi}{3}t+ \phi \right)$ . The phase difference between them at

$t=1\ s$ is:

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$\pi$
0%
$\dfrac {\pi }{2}$
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$\dfrac {\pi }{4}$
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$\dfrac {\pi}{6}$

Explanation

$t=1 sec, y_1=a\sin(\pi /2+\phi)$ .....(1)

$y_2=b\sin(2\pi/3+\phi)=b\sin(\dfrac{\pi}{2}+\dfrac{\pi}{6}+\phi)...$ (2)

From (1) and (2) it is clear that,

Phase difference

$=\dfrac{\pi}{6}$

The displacement y of a particle varies with time t as

$y = 4 \sin\omega t\ \cos \omega t$ cm where t is in seconds. The motion of the particle is

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simple harmonic with amplitude 2 cm
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simple harmonic with period $\cfrac{\pi }{\omega }$
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simple harmonic with period $\cfrac{2\pi }{\omega }$
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both a & b

Explanation

$y=4\sin\omega t\cos\omega t$

$=2\sin 2\omega t$

$A=2cm,$ Performing SHM.

A point source emits sound equally in all directions in a non-absorbing medium. Two points

$P$ and

$Q$ are at distance of

$2\ m$ and

$3\ m$ respectively from the source. The ratio of the intensities of the wave at

$P$ and

$Q$ is:

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$9:4$
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$2:3$
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$3:2$
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$4:9$

Explanation

intensity is inversely proportional to

$r^2$

$I_1:I_2=r_2^2:r_1^2=3^2:2^2=9:4$

A wave pulse on a string has the dimension shown in figure. The wave speed is

$v=1\ cm/s$ . If point

$O$ is a free end. The shape of wave at time

$t=3\ s$ is

Explanation

$Concept:$ Reflection of a pulse from a free boundary is really the superposition of two identical waves travelling in opposite directions(see the image above).

In 3 second, the real and imaginary wave will superpose each other. They both will move 3 cm towards each other. So their amplitude gets doubled (basically their enegy add up) i.e.

$2 \times 1 \ cm = 2 \ cm$

So, the correct option is

$Option \ D$

Period of a particle performing S.H.M is 4 s. It starts motion from mean position. After 2/3 s, its velocity is 6.28 cm/s. The amplitude of S.H.M is:

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(a) 8 cm
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(b) $\frac{4}{\sqrt{3}}$
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(c) $\frac{8}{\sqrt{3}}$
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(d) 4 cm

Two coherent transverse waves of amplitude 2A and A of same frequency propagated respectively along

$+ve$ and

$-ve$ x-axis are superimposed. The amplitude of the resultant wave at a distance of 5 cm from the origin,

$(given K = \cfrac{\pi}{4}cm^{-1})$ is

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$\sqrt{5A}$
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4 A
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5 A
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$\sqrt{3A}$

Two particles

$P$ and

$Q$ describe SHM of same amplitude

$a$ and frequency

$f$ along the same straight line. The maximum distance between two particles is

$a\sqrt{2}$ . The initial phase difference between the particles is:

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

Explanation

let the phase difference = z

the equation of displacement of the particle p from origin

$y=rsin(\omega t + z)$

the equation of displacement of the particle q from origin

$x=rn(\omega t)$

$2rsin(\dfrac{z}{2})= \dfrac{r}{\sqrt{2}}$

given max distance between the particles is

$\sqrt{2}r$

hence maximum phase difference is zero

Two waves of amplitudes

$A_0$ and

$xA_0$ pass through a region.If

$x>1$ ,what is the difference in the maximum and minimum resultant amplitude is :

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$(x+1)A_0$
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$(x-1)A_0$
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$2xA_0$
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$2A_0$

Explanation

Given, Waves having the amplitude

$A_0 \ and xA_0$

So, the maximum amplitude

$=A_0+xA_0$ (it is the constructive superposition),

The minimum amplitude

$=xA_0-A_0$ (it is a destructive superposition).

Therefore the difference between maximum and minimum amplitude

$=xA_0+A_0-(xA_0-A_0)=2A_0$

The amplitude of a wave disturbance propagating in the positive y-direction is given by:

$y = \dfrac{1}{1 + x^2}$ at

$t = 0s \ and \ y = \dfrac{1}{1 + (x - 1)^2}$ at

$t = 2s$ , then wave velocity is

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$1 m/s$
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$1.5 m/s$
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$0.5 m/s$
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$2 m/s$

Explanation

$2$ sec the wave moves

$1$ m

$\therefore$ Velocity

$= \dfrac{1}{2}\,\,m/s$ .

$\therefore$ Option

$C$ is correct.

The light waves from two independent monochromatic light sources are given by-

${y_1} = 2\sin {\omega t }$ and

${y_2} = 2\cos {\omega t }$ ,
then the following statement is correct

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Both the waves are coherent
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Both the waves are incoherent
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Both the waves have different time periods
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None of the above

Explanation

${y_1} = 2\sin \left( {\omega t - kx} \right)$

${y_2} = 2\cos \left( {\omega t - kx} \right) = 2\sin \left( {\omega t - kx + \frac{\pi }{2}} \right)$
So, Phase difference =

$\frac{\pi}{2}$
Hence sources are incoherent.

The displacement of a particle executing S.H.M is given by

$x=0.25sin200t$ in meters.The maximum speed of particle is :

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$50m{s^{ - 1}}$
0%
$25m{s^{ - 1}}$
0%
$200m{s^{ - 1}}$
0%
$100m{s^{ - 1}}$

Explanation

$\begin{array}{l}{V_{\max }} = Aw\\when\,\cos wt = 1\\x = 0.25\cos \left( {200t} \right)\\{V_{\max }} = 0.25 \times 200 = 50m{s^{ - 1}}\end{array}$

The displacements of two interfering light waves are

$y_1 = 4 \ sin (\omega t) \ and \ y_2 = 3 \ cos (\omega t).$ The amplitude of the resultant wave is (

$y_1$ and

$y_2$ are in CGS system)

0%
$5 cm$
0%
$7 cm$
0%
$1 cm$
0%
$zero$

Explanation

$y_{1}=A \sin{(\omega t)}$

$y_{2}=3 \sin{(\omega t)}$

Amplitude of first wave=

$y_{1}=4=A_{1}$

Amplitude of second wave=

$A_{2}=3$

Resultant=

$\sqrt{(4)^{2}+(3)^{2}}=\sqrt{16+9}=5cm$

The same progressive wave is represented by two graphs

$I$ and

$II$ . Graph

$I$ shows how the displacement

$'y'$ varies with the distance

$x$ along the wave at a given time. Graph

$II$ shoes how

$y$ varies with time

$t$ at a given point on the wave. The ratio measurements

$AB$ to

$CD$ , marked on the curves, represents:

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Wave number $k$
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Wave speed $V$
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Frequency $v$ .
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Angular frequency $\omega$

A body executing S.H.M has a maximum acceleration equal to

$48$ m

$/\sec^2$ and maximum velocity equal to

$12$ m/

$\sec$ . The amplitude of S.H.M is?

0%
$3$ m
0%
$3/32$ m
0%
$1024/9$ m
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$64/9$ m

Explanation

Form maximum velocity

$\text{ v }_{ max }=\quad A\omega ;$ =12 m/s

and from maximum acceleration

$\text{ a }_{ max }=\quad A{ \omega }^{ 2 }$ =48

$m/{ s }^{ 2 }$

where A is the amplitude of SHM.

Putting the value of

$\omega$ =

$\dfrac { 12 }{ A }$ in maximum acceleration we get ,

$\dfrac { 144 }{ { A }^{ 2 } } \times A=48\\ A=\dfrac { 144 }{ 48 } =3\quad \quad$ m

Two

$SHM$ are represented by equation

$x_{1}=4\ \sin (\omega t+37^{o})$ and

$x_{2}=5\ \cos (\omega t)$ . The phase difference between them is

0%
$37^{o}$
0%
$127^{o}$
0%
$53^{o}$
0%
$143^{o}$

Explanation

Given,

$x=A\sin (\omega t+\phi ),\,\,$

$where,\,\,$

$A=$ Amplitude

$\phi =$ Phase Angle

${{x}_{1}}=4\ \sin (\omega t+{{37}^{o}})\,\ldots \ldots \,(1)$

${{x}_{2}}=5\ \cos \text{ }\!\!~\!\!\text{ }(\omega t)=5\sin (\omega t+{{90}^{o}})\,\ldots \ldots \,(2)$

Phase angle difference is ${{90}^{o}}-{{37}^{o}}={{53}^{o}}$

A block is kept on a rough horizontal plank. The coefficient of friction between the block and the plank is 1/The plank is undergoing SHM of angular frequency 10 rad/s. The maximum amplitude of plank in which the block does not slip over the plank is (

$g = 10 \ m/s^2$ )

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$4 cm$
0%
$5 cm$
0%
$10 cm$
0%
$16 cm$

Explanation

$\textbf{Step 1 - Draw free body diagram of extreme position}$

$\textbf{Step 2 - Applying Newton's 3rd Law}$

$N = mg$

$....(1)$

$f_{max} = ma$

$....(2)$

$\because f_{max} = \mu N$

$\therefore f_{max} = \mu mg$

$....(3)$

$\textbf{Step 3 - In SHM}$

$a_{max} = \omega^{2}A$

$....(4)$

from

$(2), (3)$ and

$(4)$ , we have

$\mu mg = m\omega^{2}A$

$A = \dfrac {\mu g}{\omega^{2}}$

$....(5)$

Putting above in

$(5)$ , we get

$A = \dfrac {\dfrac {1}{2} \times 10}{10^{2}} = \dfrac {1}{20} m = \dfrac {100}{20} cm$

$A = 5\ cm$

option (B)

Two sources of waves are called coherent if:-

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Both have the same amplitude of vibrations
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Both produce waves of the same wavelength
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Both produce waves of the same wavelength having constant phase difference
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Both produce waves having the same velocity

Explanation

$\textbf{Explanation:}$ Coherent source of light are those sources which emit a light wave having the same frequency, wavelength and in the same phase or they have a constant phase difference

Hence (c) option is correct

A particle is performing simple harmonic motion bout its equilibrium position

$X_0$ , then

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$X_0$ must be zero
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velocity of particle at $X_0$ may be zero
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acceleration of particle at $X_0$ may be zero
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acceleration of particle at $X_0$ must be zero

Explanation

Aparticle is performing simple harmonic motion bout its equilibrium position

$X_0$ then

$a=\cfrac{d^2x}{dt^2}\\ \quad=\cfrac{d^2x_0}{dt^2}$

Due to its equilibrium position

$X_0$ the velocity will be uniform. hence

acceleration is

$\cfrac{V_2-V_1}{t}=1=a$

$V_2=$ Final velocity

$V_1=$ Initial velocity

$t=$ time internal

from a line source, if amplitude of a wave at a a distance r is A, then the amplitude at a distance 4r will be

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

Explanation

$\begin{array}{l} I\left( { { { intensity } } } \right) \times \frac { 1 }{ { { r^{ 2 } } } } \\ I\times { A^{ 2 } } \\ \therefore ,\, \, A\times \frac { 1 }{ r } \\ if\, \, r\to 4r,A\to \frac { A }{ 4 } \end{array}$

Two waves of the same amplitude and frequency arrive at a point simultaneously. what should the phase difference between the waves so that amplitude of the resultant wave is double(2A)

0%
$\dfrac {\pi}{2} radian$
0%
$\dfrac {2\pi}{3} radian$
0%
$\dfrac {3\pi}{4} radian$
0%
zero

The wavelength of the wave is:

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$6 \ cm$
0%
$24 \ cm$
0%
$12 \ cm$
0%
$16 \ cm$

Explanation

(Refer to Image)

A simple harmonic plane along

$x$ axis in a medium

For

$x=0$ (curve 1)

and

$x=7$ (curve 2)

The two particles are within a span of the wavelength= ?

$\lambda= \cfrac {2 \Pi}{T}$

or,

$\lambda= \cfrac {2\Pi}{\cfrac {7}{400}-\cfrac {1}{400}}=12 cm$ .

1228964_1118884_ans_d90f17bb92d84d9bb2ee43397283d98c.png

A nylon guitar string has a linear density of

$7.20\ g/m$ and is under tension of

$150\ N$ . The fixed supports are distance

$D= 90.0\ cm$ apart. The string is oscillating in the standing wave pattern shown in figure.Calculate the
(ii) The wavelength of the traveling waves whose superposition gives this standing wave.

0%
20 cm
0%
40.0 cm
0%
60.0 cm
0%
80.0 cm

A particle performing simple harmonic motion along x-axis, with

$x = 0$ as the mean position is released from rest at

$x = 2 cm$ at

$t = 0$ . The time taken by particle in crossing the position

$x = 1.6cm$ for the second time is: [Take amplitude of simple harmonic motion as

$2cm$ and its period as

$1s$ ]

0%
$\dfrac{1}{12}s$
0%
$\dfrac{11}{12}s$
0%
$\dfrac{323}{360}s$
0%
$\dfrac{23}{66}s$

A sine wave is travelling in a medium. A particular particle has zero displacement at a certain instant. the particle closest to it having zero displacement is at a distance

0%
$\lambda/4$
0%
$\lambda/3$
0%
$\lambda/2$
0%
$\lambda$

Explanation

The minima of the sine wave function occurs at 0 and after this, the next minima of the sine wave occurs at

$\pi$ . The wavelength of the sine wave function is

$2\pi$ i.e. one wave completes its full waveform in period of

$2\pi$ .

then the particle closest to it having zero displacement is at a distance

$\pi$ and is equal to

$\lambda/2$

A particle moves along the X-axis according to the equation

$x = 10{\sin ^3}(\pi t)$ . the amplitudes and frequencies of component SHMs are

0%
$amplitude 30/4, 10/4; frequencies 3/2,1/2$
0%
$amplitude 30/4 , 10/4; frequencies 1/2 , 3/2$
0%
$amplitude 10 , 10 ; frequencies 1/2 , 1/2$
0%
$amplitude 30/4 ,10 ; frequencies 3/2,2$

Explanation

Given

$x=10{ sin }^{ 3 }(\pi t)$

It is in the form

$x=\quad A{ sin }^{ 3 }(\omega t)$

where A= Amplitude. Here, A=10

We know that

$w=2\pi f$

Here

$\\ 2\pi f=\quad \pi \\ f=\dfrac { 1 }{ 2 }$

A wave of frequency

$100$ Hz is sent along a string towards a fixed end. When this wave travels back after reflection, a node is formed at a distance of

$10$ cm from the fixed end of the string. The speeds of incident(and reflected) waves are?

0%
$5$ $cms^{-1}$
0%
$10$ $cms^{-1}$
0%
$20$ $cms^{-1}$
0%
$40$ $cms^{-1}$

Explanation

Given frequency

$\nu$ =100 H.

Also distance between two successive nodes

$=>\dfrac { \lambda }{ 2 } =10\\ \lambda =20\quad cm$

And we know

$\nu =n\lambda \\ n=\frac { \nu }{ \lambda } =\dfrac { 100 }{ 20 } =5\quad cm/s$

Two periodic waves of amplitudes

$A_{1}$ and

$A_{2}$ pass through a region. If

$A_{1} > A_{2}$ , the difference in the maximum and minimum resultant amplitude possible is

0%
$2A_{1}$
0%
$2A_{2}$
0%
$A_{1}+A_{2}$
0%
$A_{1}-A_{2}$ .

The maximum particle velocity is

$8$ times the wave velocity of a progressive wave. If the amplitude of the particle is

$"a"$ . The phase difference between the two particles seperated by a distance of

$""x"$ is

0%
$\frac{x}{a}$
0%
$\frac{{8x}}{a}$
0%
$\frac{{3a}}{x}$
0%
$\frac{{3\pi x}}{a}$

The phase difference between two waves represented by

${y_1} = {10^{ - 6}}\sin \left[ {100t + \frac{x}{{50}} + 0.5} \right]m$

${y_2} = {10^{ - 6}}\cos \left[ {100t + \frac{x}{{50}}} \right]m$
where x is expressed in metres and t is expressed in seconds is approximately

0%
1.07 rad
0%
2.07 rad
0%
0.5 rad
0%
1.5 rad

Explanation

$y_{1}=10^{-6} \sin \left[100 t + \dfrac{x}{50}+ 0.5\right] m$

$y_{2}=10^{-6} \cos \left[100 t + \dfrac{x}{50}\right] m$

So,

$y^{1}=10^{-6} \cos\left[\dfrac{\pi}{2}-\left(100 t + \dfrac{x}{50}+0.5\right)\right]$

$y^{1}=10^{-6} \cos \left[\dfrac{-\pi}{2}+100 t+\dfrac{x}{50}+0.5\right]$

$\triangle \phi =\phi_{2}-\phi_{1}$

$\phi_{2}=0$ where as

$\phi_{1}=0.5-\dfrac{\pi}{2}$

So,

$\triangle \phi=\phi_{2}-\phi_{1}=\dfrac{\pi}{2}-0.5=1.07 \ rad$

option

$-A$ is correct.

The displacement of a particle in SHM is

$x =3 sin(20\pi t) +4 cos (20\pi t)$ cm.Its amplitude of oscillation is

0%
3 cm
0%
4 cm
0%
5 cm
0%
25 cm

Explanation

Both the SHM have same frequency ,with an amplitude of

$3$ and

$4$ , with a phase difference of

$90^0$

The maximum value of the equation,

$x=3\sin(20\pi t)+4\cos(20\pi t)$ is

$\sqrt{3^2+4^2}= 5\ cm$

Option

$\textbf C$ is the correct answer

Two identical sinusoidal waves each of amplitude

$5 mm$ with a phase difference of

$\dfrac{\pi}{2}$ are travelling in the same direction in a string. The amplitude of the resultant wave (in mm) is:-

0%
Zero
0%
$5 \sqrt{2}$
0%
$\dfrac{5} {\sqrt{2}}$
0%
$2.5$

Explanation

Given amplitude = 5mm

So, sinusoidal wave can be

taken as

$A = 5 sin\omega t$

Another wave has

$\dfrac{\pi}{2}$ phase diff-

$\Rightarrow B = 5 sin(\omega t+\dfrac{\pi}{2})$

$= 5 cos \omega t$

$A+B = 5(sin\omega t + cos \omega t )$

$= 5\sqrt{2}(\dfrac{1}{\sqrt{2}}sin\omega t + \dfrac{1}{\sqrt{2}} cos\omega t)$

$= 5\sqrt{2} (sin\omega t \,cos 45^{\circ}+cos \omega t\, sin45^{\circ})$

$= 5\sqrt{2} (sin(\omega t +45^{\circ}))$

$= 5\sqrt{2} sin (\omega t+ \pi/4)$

A progressive wave moves with a velocity of

$36m/s$ in a medium with a frequency of

$200Hz$ . The phase difference between two particles separated by a distance of

$1cm$ is:

0%
${40 }$
0%
$20$
0%
$\dfrac{\pi }{9}$
0%
$\dfrac{{{\pi }}}{16}$

Explanation

Solution we have,

$V = 36m/s, f = 200H_{2}$ and

$\Delta x = 1cm = 10^{-2}m$

we know that,

$v = f\lambda$

$\Rightarrow \lambda = \dfrac{v}{f} = \dfrac{36}{200} = \dfrac{9}{50}m$

Also,

$\Delta \phi = 2\pi (\dfrac{\Delta x}{\lambda })$

$\Rightarrow \Delta \phi = 2\pi \dfrac{(\dfrac{1}{100})}{\dfrac{9}{50}} \Rightarrow \Delta \phi = \dfrac{\pi }{4}$

Option - c is correct

1128666_1144635_ans_47f2fd55fabe4e98ac18da59480b0b7d.jpg

The value of phase, at maximum displacement from the mean position of a particle, in SHM is ?

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

CBSE Questions for Class 11 Engineering Physics Waves Quiz 9 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Waves Quiz 9 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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