MCQ Questions for Class 11 Engineering Physics Waves Quiz 14

The effects are produced at a given point in space by two waves described by the equations

$\displaystyle y_{1}=y_{m}\sin \omega t\: \:$ and

$\: \: y_{2}=y_{m}\sin \left ( \omega t+\phi \right )$ where

$\displaystyle y_{m}$ is the same for both the waves and

$\displaystyle \phi$ is a phase angle. Tick the correct statement among the following

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the maximum intensity that can be achieved at a point is twice the intensity of either wave and occurs if $\displaystyle \phi=0$
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the maximum intensity that can be achieved at a point is four times the intensity of either wave and occurs if $\displaystyle \phi=0$
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the maximum amplitude that can be achieved at the point its twice the amplitude of either wave and occurs at $\displaystyle \phi=0$
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When the intensity is zero the net amplitude is zero and at this point $\displaystyle \phi=\pi/4$

Explanation

${ y }_{ 1 }={ y }_{ m }\sin { wt } \\ { y }_{ 2 }={ y }_{ m }\sin { \left( wt+\phi \right) } \\ y={ y }_{ 1 }+{ y }_{ 2 }\\ y={ y }_{ m }\sin { wt } +{ y }_{ m }\sin { \left( wt+\phi \right) } \\ y={ y }_{ m }2\left[ \sin { \dfrac { (wt+wt+\phi ) }{ 2 } \cos { \dfrac { \left( wt-wt-\phi \right) }{ 2 } } } \right] \\ y=2{ y }_{ m }\cos { \left( \dfrac { \phi }{ 2 } \right) } \sin { \left( \dfrac { 2wt+\phi }{ 2 } \right) } \\ \left[ y=2{ y }_{ m }\cos { \left( \dfrac { \phi }{ 2 } \right) } \sin { \left( wt+\dfrac { \phi }{ 2 } \right) } \right] \longrightarrow$ Equation of resultant wave

$\phi =0\\ \cos { \left( \dfrac { \phi }{ 2 } \right) } =1\\ \therefore y=2{ y }_{ m }\sin { \left( wt+\dfrac { \phi }{ 2 } \right) }$

Where maximum intensity=

$2{ y }_{ m }$

Hence option (A) is correct

The motion of the particle in simple harmonic motion is given by

$x=a\sin \omega t$
If its speed is u, when the displacement is

$x_{1}$ and speed is v, when the displacement is

$x_{2}$ , show that the amplitude of the motion is
.

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$\displaystyle a=\left [ \frac{v^{3}x_{1}^{3}-u^{2}x_{2}^{2}}{v^{2}-u^{2}} \right ]^{1/2}$
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$\displaystyle a=\left [ \frac{v^{2}x_{1}^{2}-u^{2}x_{2}^{2}}{v^{2}-u^{2}} \right ]^{1/2}$
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$\displaystyle a=\left [ \frac{v^{3}x_{1}^{2}-u^{3}x_{2}^{2}}{v^{3}-u^{3}} \right ]^{1/2}$
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$\displaystyle a=\left [ \frac{v^{4}x_{1}^{2}-u^{4}x_{2}^{2}}{v^{4}-u^{4}} \right ]^{1/2}$

The following figure depicts a wave travelling in a medium. Which pair of particles are in phase

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A and D
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B and F
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C and E
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B and G

A harmonic wave is travelling on stringAt a junction with string 2, it is partly reflected and partly transmitted. The linear mass density of the second string is four times that of the first string and the boundary between the two strings is at x =If the expression for the incident wave is

$\displaystyle y_{1}=A_{1}\: \cos (k_{1}x-\omega _{1}t)$

What is the equation for the reflected wave in terms of

$\displaystyle A_{1},k_{1}$ and

$\displaystyle \omega _{1}$ ?

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$\displaystyle y =\frac{A_{1}}{4}\cos (k_{1}x+\omega _{1}t+\pi )$
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$\displaystyle y =\frac{A_{1}}{6}\cos (k_{1}x+\omega _{1}t+\pi )$
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$\displaystyle y =\frac{A_{1}}{3}\cos (k_{1}x+\omega _{1}t+\pi )$
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$\displaystyle y =\frac{A_{1}}{3}\cos (2k_{1}x+\omega _{1}t+\pi )$

Explanation

Since

$\displaystyle v=\sqrt{T/\mu }$

$\displaystyle T_{2}=T_{1}\: \: and\: \: \mu _{2}=4\mu _{1}$
we have

$\displaystyle v_{2}=\frac{v_{1}}{2}$ ...........(i)
The frequency does not change that is

$\displaystyle \omega _{1}=\omega _{2}$ ................(ii)
Also because

$\displaystyle k=\omega /v$ the wave numbers of the harmonic waves in the two strings are related by

$\displaystyle k_{2}=\frac{\omega _{2}}{v_{2}}=\frac{\omega _{1}}{v_{1}/2}=2\frac{\omega _{1}}{v_{1}}=2k_{1}$ ...................(iii)
The amplitudes are

$\displaystyle A_{1}=\left ( \frac{2v_{2}}{v_{1}+v_{2}} \right )A_{1}=\left [ \frac{2(v_{1}/2)}{v_{1}+(v_{1}/2)} \right ]A_{1}=\frac{2}{3}A_{1}$ .............(iv)
and

$\displaystyle A_{1}=\left ( \frac{v_{2}-v_{1}}{v_{1}+v_{2}} \right )A_{1}=\left [ \frac{(v_{1}/2)-v_{1}}{v_{1}+(v_{1}/2)} \right ]A_{1}=\frac{A_{1}}{3}$ .................(v)
Now with equation (ii), (iii) and (iv) the transmitted wave can be written as

$\displaystyle y_{1}=\frac{2}{3}A_{1}\cos (2k_{1}x-\omega _{1}t)$ Ans
Similarly the reflected wave can be expressed as

$\displaystyle =\frac{A_{1}}{3}\cos (k_{1}x+\omega _{1}t+\pi )$

Waves transfer

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Matter
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Particles
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Energy
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Water

The friction coefficient between the two blocks shown in the figure is

$\displaystyle \mu$ and the horizontal plane is smooth. What can be the maximum amplitude

$(A)$ if the upper block does not slip relative to the lower block ?

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$\displaystyle A=\frac{ (M+m)g}{k}$
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$\displaystyle A=\frac{\mu (M+m)g}{k}$
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$\displaystyle A=\frac{\mu (M+\mu m)g}{k}$
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$\displaystyle A=\frac{\mu (\mu M+m)g}{k}$

A particle is executing SHM

$x=3\cos{\omega t} +4\sin{\omega t}$ . Find the phase shift and amplitude.

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$50^{\circ},5$ units
0%
$37^{\circ},3.5$ units
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$53^{\circ},3.5$ units
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$37^{\circ},5$ units

Explanation

$x=3\cos { \omega t } +4\sin { \omega t }$
lets take

$x_{ 0 }\cos \phi =4$ ............

$(I)$
and

$x_{ 0 }\sin \phi =3$ ............

$(II)$
Now the equation becomes

$x=x_{ 0 }\sin \phi \cos { \omega t } +x_{ 0 }\cos \phi \sin { \omega t } \\ \Rightarrow x=x_{ 0 }\left( \sin \phi \cos { \omega t } +\cos \phi \sin { \omega t } \right) \\ \Rightarrow x=x_{ 0 }\sin { \left( \omega t+\phi \right) }$
Squaring and adding

$(I)$ and

$(II)$ we get

$\Rightarrow { x_{ 0 } }^{ 2 }(\cos ^{ 2 }{ \phi } +\sin ^{ 2 }{ \phi } )={ 4 }^{ 2 }+{ 3 }^{ 2 }=25\\ \Rightarrow x_{ 0 }=5$
dividing

$(II)$ by

$(I)$ , we get

$\tan \phi =\frac { 3 }{ 4 } \\ \Rightarrow \phi =37^{ \circ }$

$\Rightarrow x=5\sin { \left( \omega t+37^{ \circ } \right) }$ which represents a SHM of amplitude of 5 units and a phase of

$37^{ \circ }$

A person standing between the two vertical cliff produces a sound. Two successive echoes are heard at 4 s and 6 s. Calculate the distance between the cliffs :

(Speed of sound in air

$= 320 m s^{-1}$ )

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1600 m
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800 m
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400 m
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1200 m

Explanation

Distance traveled by the sound from the

$1^{st}$ cliff is calculated with the formula:
Distance traveled

$=$ velocity of sound

$\times$ time taken

That is,

$320 \times 4$

$= 1280\ m$ .

This is twice the minimum distance between a source of sound (man) and the reflector (

$1^{st}$ cliff) as it is reflected sound.
Distance traveled by the sound from the

$2^{nd}$ cliff is calculated with the formula:
Distance traveled

$=$ velocity of sound

$\times$ time taken

That is,

$320 \times 6 = 1920\ m$ .

This is twice the minimum distance between a source of sound (man) and the reflector (

$2^{nd}$ cliff) as it is reflected sound.
Hence, the distance between the 2 cliffs is given as:

$\dfrac { 1280 }{ 2 } +\dfrac { 1920 }{ 2 } =640+960=1600\ m$

Two wires of same material and area of cross section each of length 30 cm and 40 cm are stretched between two ends with tensions 10 N and 20 N respectively. The difference between the fundamental frequencies of two wires is 4.0 Hz, find the linear mass density of the wire.

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$16.4 \times 10^{-4} kg m^{-1}$
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$16.4 \times 10^{-3} kg m^{-1}$
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$6.4 \times 10^{-3} kg m^{-1}$
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$6.4 \times 10^{-5} kg m^{-1}$

If for a particle moving in SHM, there is a sudden increase of

$1$ % in restoring force just as particle passing through mean position, percentage change in amplitude will be

0%
$1$ %
0%
$2$ %
0%
$0.5$ %
0%
zero

Explanation

As force in a simple harmonic motion is directly proportional to the square of amplitude.

$F\propto { A }^{ 2 }$

$\dfrac { \triangle F }{ F } =2\dfrac { \triangle A }{ A } \\ \,\,\,\,\,\,\,\,\,\, =\dfrac { \triangle F }{ F } \times 100$

$\,\,\,\,\,\,\,\,\,\,\,=2\dfrac { \triangle A }{ A } \times 100\\$ % change in A.

% change in amplitude

$\,\,\,\,\,\,=\dfrac { 1 }{ 2 } =0.5%$

The velocity and amplitude of the component traveling waves are respectively

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50 cm/s; 0.5 mm
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50 cm/s; 5 mm
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10 cm/s; 0.5 mm
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50 cm/s; 1 mm

Explanation

The standing wave is formed by the superposition of the waves

$\displaystyle y_{1}=\frac{A}{2}\sin (\omega t-kx)$ and

$\displaystyle y_{2}=\frac{A}{2}\sin (\omega t-kx)$
The wave velocity (magnitude) of either of the waves is

$\displaystyle v=\frac{\omega }{k}=\frac{78.5s^{-1}}{1.57cm^{-1}}=50$ cm/s; Amplitude = 0.5 mm

Equations of a stationary wave and a travelling wave are

${ y }_{ 1 } = a\ sinkx\ cos \omega t$ and

${ y }_{ 2 } = a\ sin (\omega t - kx)$ . The phase difference between two points

${ x }_{ 1 }\ =\ \dfrac { \pi }{ 3k } \ and\ { x }_{ 2 }\ =\ \dfrac { 3\pi }{ 2k } \ is\ { \phi }_{ 1 }$ for the first wave and

${ \phi }_{ 2 }$ for the second wave. The ratio

$\dfrac { { \phi }_{ 1 } }{ { \phi }_{ 2 } }$ is :

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$1$
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$\dfrac{5}{6}$
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$\dfrac{3}{4}$
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$\dfrac{6}{7}$

Explanation

Phase difference between two points in a standing wave

$=n\pi$

where

$n$ is the no of nodes between two points.

Given points are

${ x }_{ 1 }=\pi /3K=60/K$

${ x }_{ 2 }=3\pi /2K=210/K$

Equation of the standing wave.

${ y }_{ 1 }=asinKxcoswt$

at node points,

$Kx=n\pi$

$x=\dfrac { n\pi }{ K }$

$(n=0,1,2,3......)$

so nodes are

$=\dfrac { \pi }{ K } ,\dfrac { 2\pi }{ K } ,\dfrac { 3\pi }{ K } .....$

$\dfrac { 180 }{ K } ,\dfrac { 360 }{ K } ,\dfrac { 540 }{ K }$

Since there is only one node in b/w phase difference

${ \phi }_{ 1 }=\pi$

for travelling wave

${ \phi }_{ 2 }=\dfrac { 2\pi }{ \lambda } \triangle x$

from the equation,

${ y }_{ 2 }=a\sin\left( wt-Kx \right)$

$K=\dfrac { 2\pi }{ \lambda }$

$\therefore \quad { \phi }_{ 2 }=K\left[ { x }_{ 2 }-{ x }_{ 1 } \right] =K\left[ \dfrac { 3\pi }{ 2K } -\dfrac { \pi }{ 3K } \right]$

$=\pi \left[ \dfrac { 3 }{ 2 } -\dfrac { 1 }{ 3 } \right] =\dfrac { 7 }{ 6 } \pi$

$\therefore \dfrac{\phi_{1}}{\phi_{2}}=\dfrac{\pi}{\dfrac{7}{6}\pi}$

$\,\,\,\,\,\,\,\,\,\,\,\,=\dfrac{6}{7}$

A horizontal stretched string, fixed at two ends, is vibrating in its

$5^{th}$ harmonic according to the equation,

$y(x, t) = (0.10\ m)\sin [(62.8\ m^{-1})x]\cos [(628s^{-1})t]$ . Assuming

$\pi = 3.14$ , the correct statement(s) is/are :

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The number of nodes is $5$
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The length of the string is $0.25\ m$
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The maximum displacement of the midpoint of the string, from its equilibrium position is $0.01\ m$
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The fundamental frequency is $100\ Hz$

Explanation

According to question we have

No of nodes

$=5$

equation of wave :

$y=\left( 0.1 \right) sin\left( 62.8x \right) cos\left( 628t \right)$

From the equation wave equation.

$K=62.8$

$\Rightarrow \dfrac { 2\pi }{ \lambda } =62.8$

$\lambda =\dfrac { 2\times 3.14 }{ 62.8 } =0.1m$

Now,

For

${ 5 }^{ th }$ Harmonic motion

$L=2\lambda +\dfrac { \lambda }{ 2 } =\dfrac { 5\lambda }{ 2 }$

$\Rightarrow \quad L=\dfrac { 5\times 0.1 }{ 2 } =0.25m$ (Length of the string)

Amplitude, i.e. maximum displacement

$=0.1m$

Fundamental frequency,

$F=\dfrac { 1 }{ 2L } \sqrt { \dfrac { T }{ \mu } } \quad OR\quad \dfrac { W/K }{ 2L }$

$=\dfrac { 628/62.8 }{ 2\times 0.25 } =20Hz$

$\therefore$ Option (A) & (B) are correct.

A wave travelling along positive x-axis is given by

$=A\sin { \left( \omega t-kx \right) }$ . If it is reflected from a rigid boundary such that

$80$ % amplitude is reflected, then equation of reflected wave is

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$y=A\sin { \left( \omega t+0.8kx \right) }$
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$y=-0.8A\sin { \left( \omega t+kx \right) }$
0%
$y=A\sin { \left( \omega t+kx \right) }$
0%
$y=0.8A\sin { \left( \omega t+kx \right) }$

Explanation

On getting reflected from a rigid boundary the wave suffers an additional phase change of

$\pi$ .
Hence, if

$\quad { y }_{ incident }=A\sin { \left( \omega t-kx \right) }$
Then,

${ y }_{ reflected }=(0.8A)\sin { \left( \omega t-k(-x)+\pi \right) } =-0.8A\sin { \left( \omega t+kx \right) }$

Wavelength of the light frequency

$100Hz$ is _________.

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$2\times {10}^{6}m$
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$4\times {10}^{6}m$
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$3\times {10}^{6}m$
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$5\times {10}^{6}m$

Explanation

Given that the light frequency is

$100\,Hz$

We know that the speed of the light is

$3\times 10^{8}\,m/s$

We have the equation,

$V=n\lambda$

where,

$V$ is the Wave velocity

$n$ is the Frequency

$\lambda$ is the Wavelength

$\implies$

$\lambda=\dfrac{V}{n}$

That is,

$\lambda=\dfrac{3\times 10^8}{100}=3 \times 10^6\,m$

Hence the wavelength is

$3\times10^6\,m$

A wave travels on a light string. The equation of the wave is

$Y = A\sin (kx - \omega t + 30^{\circ})$ . It is reflected from a heavy string tied to an end of the light string at

$x = 0$ . If

$64$ % of the incident energy is reflected the equation of the reflected wave is

0%
$Y = 0.8\ A\sin (kx - \omega t + 30^{\circ} + 180^{\circ})$
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$Y = 0.8\ A\sin (kx + \omega t + 30^{\circ} + 180^{\circ})$
0%
$Y = 0.8\ A\sin (kx + \omega t - 30^{\circ})$
0%
$Y = 0.8\ A\sin (kx + \omega t + 30^{\circ})$

Explanation

$y=A\sin\left( Kx-wt+{ 30 }^{ 0 } \right)$

Intensity of the reflected wave

$=64$ %

$=0.64$

Since

$I\alpha { A }^{ 2 }$

Amplitude of the reflected wave

$=\sqrt { I } =\sqrt { 0.64 }$

$=0.8$ of the original

i.e.

$80$ %

The reflected wave always have a phase difference of

$\pi$

Therefore equation of the reflected wave is

$\left[ y=0.8Asin\left( Kx+wt+{ 30 }^{ 0 }+{ 180 }^{ 0 } \right) \right]$

Hence Option (B) is correct.

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 y with

$y=\begin{cases} \dfrac { x }{ 4 } +1\quad \ for\quad -4 < x \le 0 \\ -x+1\quad for\quad 0 < x < 1 \\ 0 in any other case\end{cases}$
here y and x are centimeters. The linear mass density of the string is 50 g/m and it is under a tension of 5N.
The shape of the string is drawn at t = 0 and the area of the pulse enclosed by the string and the x-axis is measured. It will be equal to

0%
$2\ {cm}^{2}$
0%
$2.5\ {cm}^{2}$
0%
$4\ {cm}^{2}$
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$5\ {cm}^{2}$

The amplitude of a damped harmonic oscillator becomes halved in

$1$ minute. After three minutes the amplitude will become

$1/x$ of initial amplitude where

$x$ is:

0%
$2\times 3$
0%
${2}^{2}$
0%
${2}^{3}$
0%
$3\times {2}^{2}$

The phase difference between two waves, represented by

${ y }_{ 1 }={ 10 }^{ -6 }\sin { \left[ 100t+\left( x/50 \right) +0.5 \right]\ m }$ and

${ y }_{ 2 }={ 10 }^{ -6 }\cos { \left[ 100t+\left( x/50 \right) \right]\ m }$ . Where

$x$ is expressed in metre and

$t$ is expressed in seconds, is approximately

0%
$1.07\ radian$
0%
$2.07\ radian$
0%
$0.5\ radian$
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$1.5\ radian$

Displacement of particles in a string in x-direction and is represented by y. Account the following expression for y, those describing wave motion are.

0%
$\cos kx\sin\omega t$
0%
$k^2x^2-\omega^2 t^2$
0%
$\cos^2(kx+\omega t)$
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$\cos (kx^2-\omega^2 t^2)$

A string fixed at one end only is vibrating in its third harmonic. The wave function is

$y(x,t) = 0.02 sin(3.13x) cos(512t)$ , where y and x are in metres and t is in seconds. The nodes are formed at positions

0%
(0 m, 2 m)
0%
(0.5 m, 1.5 m)
0%
(0 m, 1.5 m)
0%
(0.5 m, 2 m)

Explanation

Node in formed only at the finest end of the string and the free end acts as an anti node.

In the third harmonics two nodes are formed:

$y\left( x,t \right) =0.02\sin { \left( 3.13x \right) \cos { \left( 512t \right) } }$

Standard equation given by

$y\left( x,t \right) =2a\sin { \left( \cfrac { 2\pi }{ \lambda } x \right) } \cos { \left( 2\pi vt \right) }$

Comparing both equation, we get

$3.13x\quad =\cfrac { 2\pi }{ \lambda } x\\ or,\quad \lambda =\cfrac { 2\lambda }{ 3.13 } \\ \quad \quad \quad \quad =2 m (approx)$

The nodes are formed at $\cfrac { \lambda }{ 4 } =0.5$ from origin and at $\cfrac { 3\lambda }{ 4 } =1.5$ from origin.

A wave

$10\sin{(ax+bt)}$ is reflected from dense medium at an origin. If 81% of energy is reflected then the equation of reflected wave is

0%
$y=-8.1\sin{(ax-bt)}$
0%
$y=-8.1\sin{(ax+bt)}$
0%
$y=-9\sin{(ax-bt)}$
0%
$y=-10\sin{(ax-bt)}$

Explanation

Wave equ before reaction :

$10sin(ax+bt)$

Intensity of the reflected wave

$=81$ %

$=0.81$ of the original.

Since

$I\alpha { A }^{ 2 }$

Amplitude of reflected wave

$=\sqrt { I } =\sqrt { 0.81 } =0.9$ of the original

$A$ .

$\therefore \quad { A }^{ 1 }=0.9A$

${ A }^{ 1 }=0.9\times 10=9$

Equ of reflected wave :

$\left[ y=-9\sin\left( ax-bt \right) \right]$

$\therefore$ Option (C) is correct.

A tray of mass M =

$10$ kg is supported on two identical springs, each of spring constant k, as shown in figure. When the tray is depressed a little and released, it executes simple harmonic motion of period

$1.5$ s. When a block of mass m is placed on the tray, the period of oscillation becomes

$3$ s.The value of m is

0%
$10$ kg
0%
$20$ kg
0%
$30$ kg
0%
$40$ kg

Explanation

$T_1=2\pi \sqrt{\cfrac{M}{2K}}$

$T_2=2\pi \sqrt{\cfrac{M+m}{2K}}$

$\Longrightarrow \cfrac{T_1}{T_2}=\cfrac{\sqrt{M}}{\sqrt{M+m}}$

$\Longrightarrow \cfrac{1.5}{3}=\sqrt{\cfrac{10}{10+m}}$

$\Longrightarrow \cfrac{1}{4}=\cfrac{10}{m+10}$

$\Longrightarrow m=30kg.$

Waves travelling in same medium having equations:

$y_1=A\sin(\alpha t-\beta x)$ and

$y_2=A\cos [\alpha t+\beta x-(\pi /4)]$ have different.

0%
Speeds
0%
Direction
0%
Wavelength
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Frequencies

A wave frequency

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

$10cm$ from the fixed end. The speed of the wave (incident and reflected) is

0%
$5m/s$
0%
$10m/s$
0%
$20m/s$
0%
$40m/s$

Explanation

Frequency of the wave

$=100{ H }_{ 3 }$

$\therefore$ Time period

$=\dfrac { 1 }{ f } =\dfrac { 1 }{ 100 } =0.01sec$

wavelength of the wave

$=2\times 10cm=20cm$

$\lambda =0.2m$

Speed of the wave

${ V }_{ w }=\dfrac { w }{ k } =\dfrac { 2\pi /T }{ 2\pi /\lambda } =\dfrac { \lambda }{ T } =\dfrac { 0.20 }{ 0.01 } \quad \left[ { V }_{ w }=20m/s \right]$

$\therefore$ Option (C) is correct.

970150_1027847_ans_e7fcfb4fb56d4e7faf23ab1a652cf03d.jpg

A horizontal spring-block system of mass 1 kg executes SHM of amplitude 10 cm. When the block is passing through its equilibrium position another mass of 1 kg is put on it and the two move together:

0%
amplitude will remain unchanged
0%
amplitude will become 5 V2 cm
0%
the frequency of oscillations will remain same
0%
the frequency of oscillations will decrease

The equation of a plane progressive wave is

$y = 0.02 \sin 8 \pi[t - \dfrac{x}{20}].$ When it is reflected at a rarer medium, its amplitude becomes 75% of its previous value. The equation of the reflected wave is

0%
$y = 0.02 \sin 8 \pi [t - \dfrac{x}{20}]$
0%
$y = 0.02 \sin 8 \pi [t + \dfrac{x}{20}]$
0%
$y = 0.15 \sin 8 \pi [t + \dfrac{x}{20}]$
0%
$y = 0.15 \sin 8 \pi [t - \dfrac{x}{20}]$

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
(iii) The frequency of the traveling waves whose superposition gives this standing wave.

0%
$\frac{1000}{\sqrt[3]{3}}Hz$
0%
$\frac{1250}{\sqrt[3]{3}}Hz$
0%
$\frac{1500}{\sqrt[3]{3}}Hz$
0%
$\frac{1750}{\sqrt[3]{3}}Hz$

A wave represented by

$y = 100 \sin(ax + bt)$ is reflected from a dense plane at the origin. If

$36$ % of energy is lost and rest of the energy is reflected then the equation of the reflected wave will be:-

0%
$y = -80 \sin(ax + bt)$
0%
$y = -1 \sin(ax + bt)$
0%
$y = -8.1 \sin(ax - bt)$
0%
$y = -10 \sin(ax - bt)$

Two waves of equal frequencies have their amplitudes in the ratio of 3 :They are superimposed on each other. Calculate the ratio of maximum and minimum intensities of the resultant wave.

0%
16:1
0%
15:1
0%
1:16
0%
1:15

If two waves, each of intensity

${I}_{0}$ , having the same frequency but differing by a constant phase angle of

${60}^{o}$ , superpose at a certain point in space, then the intensity of resultant wave is:

0%
$2{I}_{0}$
0%
$\sqrt{3}{I}_{0}$
0%
$3{I}_{0}$
0%
$4{I}_{0}$

A particle executes SHM with a time period of

$16$ s. At time

$t=2s$ , the particle crosses the mean position while at

$t=4s$ , its velocity is

$4ms^{-1}$ . The amplitude of motion in metre is?

0%
$\sqrt{2}\pi$
0%
$16\sqrt{2}\pi$
0%
$32\sqrt{2}/\pi$
0%
$4/\pi$

Explanation

$x=\alpha sin\left( \dfrac { 2\pi }{ T } t+\phi \right)$

$t=2s,x=0,T=16s$

$0=\alpha sin\left( \dfrac { \pi }{ 4 } +\phi \right)$

$\phi =-\dfrac { \pi }{ 4 }$

equ of shm is

$x=asin\left( \dfrac { 2\pi }{ T } t-\frac { \pi }{ 4 } \right)$

$t=4s,v=4m/s$

$v=\dfrac { dx }{ dt } =a\times \dfrac { 2\pi }{ t } cos\left( \dfrac { 2\pi }{ T } t-\dfrac { \pi }{ 4 } \right)$

$4=\alpha \times \dfrac { 2\pi }{ 16 } cos\left( \dfrac { \pi }{ 2 } -\dfrac { \pi }{ 4 } \right)$

$4=\alpha \times \dfrac { 2\pi }{ 16 } \times \dfrac { \pi }{ 8 } \times \dfrac { 1 }{ \sqrt { 2 } }$

$\alpha =\dfrac { 32\sqrt { 2 } }{ \pi }$

A particle is executing SHM of amplitude

$A$ , about the mean position

$x = 0$ . Which of the following cannot be a possible phase difference between the positions of the particle at

$x = +A/2$ and

$x = -A/\sqrt {2}$ .

0%
$75^{\circ}$
0%
$165^{\circ}$
0%
$135^{\circ}$
0%
$195^{\circ}$

Two particles execute

$S.H.M.$ along the same line at the same frequency. They move in opposite direction at the mean position. The phase difference will be:

0%
$1\pi$
0%
$2\pi/3$
0%
$\pi$
0%
$\pi/2$

Two waves are represented by

$x_1 = A \sin \left( \omega t + \dfrac{\pi}{6} \right)$ and

$x_2 = A \cos \omega t$ then the phase difference between them is :

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

Explanation

The general equation of the waves given is

${x_1} = A\sin \left( {\omega t + \dfrac{\pi }{6}} \right)$

${x_2} = A\cos \left( {\omega t} \right)$

Then ${x_1}$ can be written as

${x_1} = A\cos \left( {\dfrac{\pi }{2} - \omega t - \dfrac{\pi }{6}} \right)$

On comparing and solving we get the phase difference is $\dfrac{\pi }{3}$

Equation of a progressive wave is given by

$y = 0.2 \cos \pi \Bigg \lgroup 0.04t + .02x - \frac{\pi}{6} \Bigg \rgroup$
The distance is expressed in cm and time in second. What will be the minimum distance between two particles having the phase difference of

$\pi$ /2

0%
4 cm
0%
8 cm
0%
25 cm
0%
12.5 cm

Two waves are propagating to the point p along a straight line produced by two sources A and B of ahead by

$\pi /3$ than that of B and the distance AP is greater than BP by 50 cm. Then the resultant amplitude at the point P will be, if the wavelength is 1 meter

0%
$2a$
0%
$a\sqrt{3}$
0%
$a\sqrt{2}$
0%
$a$

Maximum acceleration of an object in simple harmonic motion is 24

$m/s^2$ and maximum velocity is 16 m/sec. The amplitude of object is

0%
$\dfrac { 32 } { 3 } \mathrm { m }$
0%
$16\ m$
0%
$\dfrac { 2 } { 3 } \mathrm { m }$
0%
$\dfrac { 3 } { 2 } \mathrm { m }$

The time taken by block-bullet system to move from

$y = \dfrac { m g } { k }$ (initial equilibrium position) to

$y = 0$ (natural length of spring) is (

$A$ represents the amplitude of motion).

0%
$\sqrt { \dfrac { 4 m } { 3 k } } \left[ \cos ^ { - 1 } \left( \dfrac { m g } { 3 k A } \right) - \cos ^ { - 1 } \left( \dfrac { 4 m g } { 3 k A } \right) \right]$
0%
$\sqrt { \dfrac { 3 k } { 4 m } } \left[ \cos ^ { - 1 } \left( \dfrac { m g } { 3 k A } \right) - \cos ^ { - 1 } \left( \dfrac { 4 m g } { 3 k A } \right) \right]$
0%
$\sqrt { \dfrac { 4 \mathrm { m } } { 6 k } } \left[ \sin ^ { - 1 } \left( \dfrac { 4 \mathrm { mg } } { 3 k A } \right) - \sin ^ { - 1 } \left( \dfrac { \mathrm { mg } } { 3 k A } \right) \right]$
0%
$None\ of\ the\ above$

Explanation

Initially in equilibrium let the elongation is spring be

$y_{0}$ then

$mg=ky_{0}$

$y_{0}=\dfrac{mg}{k}$
As the bullet strikes the block with velocity

$v_{0}$ and gets embedded into it, the velocity of the combined mass can be computed by using the principle of moment conservation.

$\dfrac{m}{3}v_{0}=\dfrac{4m}{3}v\Rightarrow v=\dfrac{v_{0}}{4}$
Let new mean position is at distance

$y$ from the origin, then

$ky=\dfrac{4m}{3}g\Rightarrow y=\dfrac{4 mg}{3k}$

Now , the block executes

$SHM$ about mean position defined by

$y=4 mg/3k$ with time period

$T=2\pi \sqrt{4m/3k}$ . At

$t=0$ , the combined mass is at a displacement of

$(y-y_{0})$ from mean position and is moving with velocity

$v$ , then by using

$v=\omega \sqrt{A^{2}-x^{2}}$ , we can find the amplitude of motion.

$\left(\dfrac{v_{0}}{4}\right)^{2}=\dfrac{3k}{4m}[A^{2}-(y-y_{0})^{2}]=\dfrac{3k}{4m}\left[A^{2}-\left(\dfrac{mg}{3k}\right)^{2}\right]$

$\Rightarrow A=\sqrt{\dfrac{mv_{0}^{2}}{12k}+\left(\dfrac{mg}{3k}\right)^{2}}$
To compute the time taken by the combined mass from

$y=mg/k$ to

$y=0$ , we can either go for equation method or circuit motion projection method.

Required time,

$t=\dfrac{\theta}{\omega}=\dfrac{\alpha-\beta}{\omega}$

$\cos a=\dfrac{y-y_{0}}{A}=\left(\dfrac{\dfrac{4mg}{3k}-\dfrac{4mg}{k}}{A}\right)=\dfrac{mg}{3kA}$

$\cos\beta=\dfrac{y}{A}=\dfrac{4mg}{3kA}$

So,

$t=\dfrac{\cos\left(\dfrac{mg}{3kA}\right)-\cos^{-1}\left(\dfrac{4mg}{3kA}\right)}{\omega}$

$t=\sqrt{\dfrac{4m}{3k}}\left[\cos^{-1}\left(\dfrac{mg}{3kA}\right)-\cos^{-1}\left(\dfrac{4mg}{3kA}\right)\right]$

1623087_1190654_ans_b3a727385d4b474dad6548997ca52dce.PNG

A person observe two points on a string as a travelling wave passes them. The points are at

$x _ { 1 } = 0$ and

$x_2 = 1m$ . The transverse motions of the two points are found to be as follows:

$y _ { 1 } = 0.2 \sin 3 \pi t$

$y _ { 2 } = 0.2 \sin ( 3 \pi t + \pi/8 )$ What is the frequency in

$Hz$ ?

0%
$1.5 Hz$
0%
$3 Hz$
0%
$4.5 Hz$
0%
$1 Hz$

A plane wave

$y=a sin(bx+ct)$ is incident on a surface. Equation of the reflected wave is

$y'=a'sin(ct-bx)$ Which of the following statement is not correct?

0%
The wave is incident on the surface normally.
0%
Reflecting surface is y-z plane.
0%
Medium, in which incident wave is travelling , is denser than the other medium.
0%
a' cannot be greater than a.

A particle starts from the origin, goes along

$X-$ axis to the point

$(20\ m,\ 0)$ and then returns along the same line to the point

$(-20m,\ 0)$ . The distance and displacement of the particle during the trip are

0%
$40m$ , $0$
0%
$40m$ , $20m$
0%
$40m$ , $-20m$
0%
$60m$ , $20m$

A particle executes simple harmonic motion with a time period of 16 s. At time t=2s, the particle crosses the mean position. Its velocity is

$4 ms^{-1}$ when t=4s. The amplitude of motion is

0%
$\sqrt{2}\, \pi m$
0%
$16 \sqrt{2}\, \pi m$
0%
$24 \sqrt{2}\, \pi m$
0%
$\dfrac{32\sqrt{2}}{\pi}m$
0%
$\dfrac{4}{\pi}m$

Two simple harmonic motions are represented by the equation

$y_1=10sin(4\pi t+\pi/4)$ and

$y_2=5(sin\ 3\pi t+\sqrt{3}cos\ 3\pi t)$ . Their amplitudes are in the ratio

0%
$1:1$
0%
$2:1$
0%
$2:\sqrt{3}$
0%
$\sqrt{3}:2$

The distance between consecutive maxima and minima is given by

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

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 :

0%
$\sqrt{ \dfrac { { a+b } }{ ab } }$
0%
$\dfrac { \sqrt { a } +\sqrt { b } }{ ab }$
0%
$\dfrac { \sqrt { a } -\sqrt { b } }{ ab }$
0%
$\dfrac { a+b }{ ab }$

A body is performing linear SHM, If the displacement, acceleration and the corresponding velocity of the body are y, 'a' and v respectievly, which of the following graphs is/are correct?

A sound wave is traveling towards right and its s-t graph is as show for x=0.
What will be density vs x graph at t=T/4:-

A wave pulse is given by the equation

$y=f(x,t)=A exp(-B(x-vt)^{2})$ . Given

$A=1.0m.B=1.0m^{-2}$ and

$v=+2.0m/s$ . which of the following graph shows the correct wave profile at the instant

$t=1s$ ?

Two waves of amplitudes 4a and 2a have a phase different of

$\pi$ between them. The resultant intensity will be :-

0%
$4a^{2}$
0%
$2a^{}$
0%
$a^{2}$
0%
$16a^{2}$

CBSE Questions for Class 11 Engineering Physics Waves Quiz 14 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Waves Quiz 14 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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