MCQ Questions for Class 11 Engineering Physics Waves Quiz 5

$P$ is the junction of two wires

$A$ and

$B$ .

$B$ is made of steel and is thicker while

$A$ is made of aluminium and is thinner as shown. If a wave pulse as shown in the figure approaches

$P$ , the reflected and transmitted waves from

$P$ are respectively :

Explanation

When a wave incidents on a denser medium from a rarer medium, it goes through

$\pi$ phase change for the reflected beam and

$0$ phase change for the transmitted beam.

Equations of motion in the same direction is given by

$y_1 = A\sin(\omega t - kx)$ ,

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

0%
$\sqrt2\ A\cos\theta$
0%
$2A\cos\theta$
0%
$\sqrt2A\cos\dfrac{\theta}{2}$
0%
$2A\cos\dfrac{\theta}{2}$

Explanation

Amplitude of the resultant wave is

$A_R=\sqrt{A^2_1+A^2_2+2A_1A_2\cos(\theta)}$ .
Here,

$A_1=A_2=A$ , so

$A_R=A\sqrt{2(1+\cos(\theta))}=2A\cos(\theta/2)$ .

Two waves of equal amplitude

$x_0$ and equal frequency travel in the same direction in a medium. The amplitude of the resultant wave is

0%
$0$
0%
$x_0$
0%
$2x_0$
0%
Between $0$ and $2x_0$

Explanation

The amplitude of the resultant wave is

$x'_0=\sqrt{x^2_{01}+x^2_{02}+2x_{01}x_{02}\cos(\phi)}$ , here the angle between the two propagating waves is

$\phi$ .

$x'_{0}$ is maximum when

$\phi = 0^o$ and it becomes

$x'_{0}=x_{01}+x_{02}=2x_0$ for

$x_{01}=x_{02}=x_0$ .

$x'_{0}$ is minimum when

$\phi = 180^o$ and it becomes

$x'_{0}=x_{01}-x_{02}=0$ for

$x_{01}=x_{02}=x_0$ .
The amplitude of the resultant wave is between 0 and 2

$x_0$ .

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is correct but Reason is incorrect
0%
Assertion is incorrect but Reason is correct

Two waves of same frequency but amplitudes equal to

$a$ and

$2a$ travelling in the same direction superimpose out of phase. The resultant amplitude will be

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

Explanation

The amplitude of the resultant wave is

$x'_0=\sqrt{a^2+ (2a)^2+2 a \times 2a\cos(\phi)}$ , here the angle between the two propagating waves is

$\phi=\pi$ .
Hence

$x'_0=\sqrt{a^2+ (2a)^2+2 a \times 2a\cos(\pi)}=\sqrt{a^2+ (2a)^2-2 a \times 2a}=a$

A wave of frequency

$400\space Hz$ has a velocity of

$320\space ms^{-1}$ . The distance between the particles differing in phase by

$90^{\small\circ}$ is

0%
$80\space cm$
0%
$60\space cm$
0%
$40\space cm$
0%
$20\space cm$

Explanation

Let,
Frequency of wave is, $f = 400 Hz$
Velocity of wave is, $v = 320 ms^{-1}$
The phase difference is, $\phi = 90^\circ = \dfrac{\pi}{2}$
The wavelength of wave is, $\lambda$
$\lambda = \dfrac{v}{f}$
$\lambda = \dfrac{320}{400}$
$\lambda = 0.8 m$
Now, the phase difference between two particles in a wave is
$\dfrac{\phi}{2\pi}= \dfrac{x}{\lambda}$
where, x is the distance between two particles in a wave.
$x = \dfrac{\lambda \phi}{2 \pi}$
$x = \dfrac{0.8 (\dfrac{\pi}{2})}{2 \pi}$
$x = \dfrac{0.8}{4} = 0.2 m$
$x = 20 cm$

Two sound waves are represented by the following equations

$\quad y_1 = 10\sin(3\pi - 0.03x)$ and

$\quad y_2 = 5\sin(3\pi t - 0.03x) + 5\sqrt3\cos(3\pi t - 0.03 x)$
Then, the ratio of their amplitudes is given by

0%
$1:1$
0%
$1:2$
0%
$2:1$
0%
$2:5$

Explanation

mistake in the question .
if it is

$5 \sqrt 3$ , then the below equation is a combination of two waves with pase difference of

$\frac {\pi} {2}.$
hence its amplitude is

$\sqrt {{5}^{2} + {(5\sqrt3)}^{2}}$ = 10 . hence ratio is 1:1

Two wires with different densities are joined at

$x = 0$ . An incident wave

$Y_i = A_i\sin(\omega t -k_1x)$ travelling from left to right is partially reflected and partially transmitted

$(Y_t = A_t\sin(\omega t -k_2x))$ at

$x = 0$ . If the amplitude of transmitted wave is

$A_t$ then

$^{\Large A_t}/_{\Large A_i}$ is

0%
$\displaystyle\frac{2k_1}{k_1+k_2}$
0%
$\displaystyle\frac{2k_2}{k_1+k_2}$
0%
$\displaystyle\frac{k_1-k_2}{k_1+k_2}$
0%
$\displaystyle\frac{\sqrt{k_1}-\sqrt{k_2}}{\sqrt{k_1}+\sqrt{k_2}}$

Explanation

Let the equation of reflected wave be

$A_rsin(\omega t+k_1x)$

The sign is reversed to include the reversed direction of the reflected wave.

From continuity of the wave at the interface

$x=0$ , we get,

$A_i+A_r=A_t$

Also the first derivative of the wave are continuous at the boundary, hence,

$k_1(A_i-A_r)=k_2A_t$

$\implies A_i-A_r=\dfrac{k_2}{k_1}A_t$

Rearranging the two equations gives,

$\dfrac{A_t}{A_i}=\dfrac{2k_1}{k_1+k_2}$

If the frequency of a sound wave is doubled then the velocity of sound will be

0%
zero
0%
half
0%
double
0%
unchanged

A boat at anchor is rocked by a wave whose crests are

$100\space m$ apart and whose velocity is

$25\space ms^{-1}$ .how often does the crest reach the boat?

0%
$2500\space s$
0%
$1500\space s$
0%
$4.0\space s$
0%
$0.25\space s$

Explanation

Distance between consecutive crests is the wavelength of the wave.

Here given is

$\lambda=100m$

$v=\lambda\nu$

$\implies \nu=\dfrac{v}{\lambda}$

$\implies \nu=\dfrac{100m}{25m/s}=4s$

For a pulse moving in a heavy string the junctions of the string behaves as a

0%
perfectly rigid end
0%
free end
0%
partially rigid end
0%
rigid end

The wavelength of a sound wave is reduced by 50%. Then the percentage change in its frequency will be

0%
100%
0%
200%
0%
400%
0%
800%

Explanation

The wavelength of sound wave is

$\lambda = \dfrac{v}{\nu}$

$\nu = \dfrac{v}{\lambda}$

where, v is the velocity of sound waves and

$\nu$ is the frequency of sound waves.

When the wavelength of a sound wave is reduced by 50%

$\lambda - \dfrac{50}{100}\lambda = \dfrac{v}{\nu'}$

$\dfrac{1}{2}\lambda = \dfrac{v}{\nu'}$

$\nu' = \dfrac{2v}{\lambda}$

The percent change in frequency is

$\dfrac{\nu - \nu'}{\nu} \times 100 = \dfrac{\dfrac{2v}{\lambda} - \dfrac{v}{\lambda}}{\dfrac{v}{\lambda}} \times 100$

$\Rightarrow (2 - 1) \times 100 = 100$ %

A rod

$70\space cm$ long is clamped from middle. The velocity of sound in the material of the rod is

$3500\space ms^{-1}$ . The frequency of fundamental note produced by it is :

0%
$3500\space Hz$
0%
$2500\space Hz$
0%
$1250\space Hz$
0%
$700\space Hz$

Explanation

SInce rod is clamped at middle, therefore only nodes can form at that point, and at free end only antinode can form, therefore for fundamental mode of frequency,

$\lambda/4=L/2$

$\lambda=2L=2\times 70 cm=1.4 m$

$v=3500\ ms^{-1}$ is given,

We know,

$v=f \times \lambda$ or

$f = \dfrac{v}{\lambda}=3500/1.4=2500 Hz$

Option "B" is correct.

A sonometer wire,

$100\ \text{cm}$ in length has a fundamental frequency of

$330\ \text{Hz}$ . The velocity of propagation of transverse waves along this wire is :

0%
$330\ \text{ms}^{-1}$
0%
$660\ \text{ms}^{-1}$
0%
$115\ \text{ms}^{-1}$
0%
$990\ \text{ms}^{-1}$

Explanation

For the fundamental frequency

$(f_0=330 Hz),\ \lambda=2 L=200\ \text{cm} = 2\ \text{m}$

then from the formula,

$v=f_0 \times \lambda= 330 \times 2.0= 660 \text{ms}^{-1}$

If the wavelength of a wave is decreased by

$20\mbox{%}$ then its frequency will become :

0%
$20\mbox{%}\space less$
0%
$25\mbox{%}\space more$
0%
$20\mbox{%}\space more$
0%
$25\mbox{%}\space less$

Explanation

Velocity of sound in the air does not change with frequency it means that the product of wavelength and frequency remains constant for different sounds.

$v=f_1 \times \lambda_1 = f_2 \times \lambda_2$

if,

$\lambda_2= 0.8 \lambda_1$ ,

$f_2 = f_1 \dfrac{\lambda_1}{ \lambda_2}= f_1 (1/0.8)=1.25 \lambda_1$

option "B" is correct.

A wave represented by

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

$36\mbox{%}$ 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%
$y = -8.1\sin(ax+bt)$
0%
$y = -10\sin(ax+bt)$
0%
$y = -8.2\sin(ax+bt)$

Explanation

Remaining energy of the reflected beam is

$(100-36)\%=64\%$ .
So intensity of the reflected beam can be given by,

$I'=64\% \text{ of } I_0$

or,

$I'=\dfrac{64}{100}A^2=\dfrac{64}{100}(100)^2=6400$ , where A is the amplitude of the incident wave.
Amplitude of the reflected beam A' can now be derived as

$I'=A'^2\Rightarrow A' =\sqrt{I'}=80$ .
Considering the change in phase due to reflection from a denser medium, we get the equation of the reflected beam as :

$y=80\sin(ax+bt+\pi)=-80\sin(ax+bt)$ .

An ultrasonic scanner is used in a hospital to detect tumours in tissue. The working frequency of the scanner is 4.2 mega Hz. The velocity of sound in the tissue is

$1.7 kms^{-1}$ . The wavelength of sound in the tissue is nearest to

0%
$\displaystyle 4 \times 10^{-3}$ m
0%
$\displaystyle 8 \times 10^{-3}$ m
0%
$\displaystyle 4 \times 10^{-4}$ m
0%
$\displaystyle 8 \times 10^{-4}$ m

Explanation

Let,

The working frequency of the scanner is

$\nu = 4.2 MHz = 4.2 \times 10^6 Hz$ .

The velocity of sound in the tissue is

$v = 1.7 kms^{-1} = 1.7 \times 10^3 ms^{-1}$

the wavelength of sound is given by

$\lambda = \dfrac{v}{\nu}$

$\lambda = \dfrac{1.7 \times 10^3}{4.2 \times 10^6}$

$\lambda = 0.404 \times 10^{-3}$

$\lambda \approx 4 \times 10^{-4} m$

In a Kundt's tube experiment, the heaps of lycopodium powder are collected at

$20\ cm$ separations. The frequency of tuning fork used is

0%
$660\ Hz$
0%
$825\ Hz$
0%
$775\ Hz$
0%
$915\ Hz$

Explanation

The in kundt's tube powder is collected where nodes are formed. The separation between nodes is given as

$d=\dfrac{\lambda}{2} =20 cm$

$\lambda=40 cm$

$f=\dfrac{c}{\lambda}$

$=\dfrac{330}{.4} =825 \ Hz$

The equation of a plane progressive wave is

$y=0.9 sin 4\pi[t-\dfrac{x}{2}]$ . When it is reflected at a rigid support, its amplitude becomes

$\dfrac{2}{3}$ of its previous value. The equation of the reflected wave is

0%
$y=0.6 sin 4\pi[t+\dfrac{x}{2}]$
0%
$y=-0.6 sin 4\pi[t+\dfrac{x}{2}]$
0%
$y=-0.9 sin 8\pi[t-\dfrac{x}{2}]$
0%
$y=-0.9 sin 4\pi[t+\dfrac{x}{2}]$

Explanation

Amplitude of reflected wave

$=0.9 \times \dfrac{2}{3}=0.6$
It would travel along negative direction of x-axis , and on reflection at a rigid support, there occurs a phase change of

$\pi$ .
Hence, the wave equation becomes:

$y = 0.6 \sin 4\pi(t+\dfrac{x}{2}+\pi)$

$y =- 0.6 \sin 4\pi(t+\dfrac{x}{2})$

A series of ocean waves, each 5.0 m from crest to crest, moving past the observer at a rate of 2 waves per second. What is the velocity of ocean waves?

0%
2.5 m/s
0%
5.0 m/s
0%
8.0 m/s
0%
10.0 m/s

Explanation

Here wavelength of the waves is given as

$\lambda = 5.0$ m.

Frequency

$\nu =2$

Wave velocity

$v = \nu \lambda= 10$ m/s.

If two waves of same frequency and same amplitude, on superposition, produce a resultant disturbance of the same amplitude, the wave differ in phase by

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

Explanation

We know that:

$R^2= a^2+b^2+2 ab cos\phi$

$\therefore a^2=a^2+a^2+2a^2 cos\phi$

$\Rightarrow cos\phi =-\dfrac{1}{2}$

$\Rightarrow \phi=\dfrac{2\pi}{3}$

The displacement

$y$ in centimeters is given in terms of time

$t$ in second by the equation:

$y=3\sin 3.14t+4\cos 3.14 t$ , then the amplitude of SHM is

0%
3 cm
0%
4 cm
0%
5 cm
0%
7 cm

Explanation

When the displacement of a SHM is

$y=a\sin wt+b\cos wt$ , the amplitude of the SHM will be

$A=\sqrt{a^2+b^2}$

Here,

$a=3, b=4$ so amplitude

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

The equation of displacement of a harmonic oscillator is

$x=3\sin{\omega t}+4\cos{\omega t}$ . The amplitude of the particles will be

0%
1
0%
5
0%
7
0%
12

Explanation

In the equation of superposition of SHMs or Harmonic oscillator,

$x=asin\omega t+bsin\omega t$ , the resultant amplitude is given by

$A=\sqrt{a^{2}+b^{2}}$ . Substituting a=3 and b=4, we get final answer as Option B, i.e. 5

If the frequency of a sound wave is increased by 25%, then the change in its wavelength will be

0%
25% decrease
0%
20% decrease
0%
20% increase
0%
25% increase

Explanation

The frequency of wave is given by

$\nu = \dfrac{v}{\lambda}$

$\lambda = \dfrac{v}{\nu}$

When the frequency of a sound wave is increased by 25%, then the new

wavelength is

$\lambda' = \dfrac{v}{\nu+\dfrac{25}{100}\nu}$

$\lambda' = \dfrac{v}{\nu+\dfrac{1}{4}\nu}$

$\lambda' = \dfrac{v}{\dfrac{5}{4}\nu}$

$\lambda' = \dfrac{4v}{5\nu}$

Hence, the percent change in wavelength is

$\dfrac{\lambda - \lambda'}{\lambda} \times 100 = \dfrac{\dfrac{v}{\nu} - \dfrac{4v}{5\nu}}{\dfrac{v}{\nu}} \times 100$

$\Rightarrow -\dfrac{1}{4} \times 100 = -20$ %.

Hence, wavelength decreases by 20%

A sound wave of frequency

$500\ \text{Hz}$ covers a distance of

$1000\ \text{m}$ in

$5\ \text{s}$ between points

$\displaystyle x$ and

$\displaystyle y$ . Then the number of waves between

$\displaystyle x$ and

$\displaystyle y$ are

0%
5000
0%
2500
0%
100
0%
500

Explanation

$\nu= 500\ \text{Hz}$

$x= 1000$

$t= 5\ \text{s}$

velocity of sound

$v=\dfrac{x}{t} = 200\ \text{m/s}$

But

$v= \nu \lambda$

Hence,

$\lambda = \dfrac{2}{5}\ \text{m/s}$

Now

$n\lambda = x\ \text{or}\ n = 2500$

Ans: B

Interference event is observed

0%
only in transverse waves
0%
only in longitudinal waves
0%
in both types of waves
0%
none

Two particles are executing

$S.H.M$ 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 amplitude. What is the phase difference between them?

0%
$5\pi/6$
0%
$2\pi/3$
0%
$\pi/3$
0%
$\pi/6$

Explanation

$\quad y = a\sin(\omega t + \phi); \mbox{ when } y = a/2 \\ \quad then\space \displaystyle\frac{a}{2} = a\sin(\omega t + \phi)$

$\quad or \space \sin(\omega t + \phi) = \displaystyle\frac{1}{2} = \sin\displaystyle\frac{\pi}{6} \space or \space \sin\displaystyle\frac{5\pi}{6}$

So phase of two particles is

$\pi/6$ and

$5\pi/6$ radians.

Hence phase difference

$= (5\pi/6) - \pi/6 = 2\pi/3$

The equation

$y=A sin^2(kx-\omega t)$ represents a wave with

0%
amplitude A frequency $\omega/2\pi$
0%
amplitude A /2 frequency $\omega/\pi$
0%
amplitude 2A frequency $\omega/4\pi$
0%
It does not represent a wave motion

Explanation

$y=A\sin^2(kx-\omega t) = \dfrac{A}{2}\{1-\cos(2kx-2\omega t)\}$
Comparing the above equation with standard progressive wave equation

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

Frequency

$\nu'=\dfrac{\omega'}{2\pi}= \dfrac{2\omega}{2\pi}=\dfrac{\omega}{\pi}$ and amplitude

$A'=A/2$
Ans: B

The phenomenon of interference is shown by

0%
longitudinal mechanical waves only
0%
transverse mechanical waves only
0%
non-mechanical transverse waves only
0%
All of the above

Statement 1: A transverse wave are produced in a very long string fixed at one end. Only progressive wave is observed near the free end.
Statement 2: Energy of reflected wave does not reach the free end

0%
Statement-1 is false, Statement -2 is true
0%
Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1
0%
Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for statement-1
0%
Statement-1 is true, Statement-2 is false

The frequency of light of wave length

$5000\ {\buildrel_{\circ}\over {A}}$ is:

0%
$1.5 \times 10^3\ Hz$
0%
$6 \times 10^8\ Hz$
0%
$6 \times 10^{14}\ Hz$
0%
$7.5 \times 10^{15}\ Hz$

Explanation

Given,

Wavelength,

$\lambda=5000\ {\buildrel_{\circ}\over {A}}$

$\lambda=5000\times 10^{-10}\ m$

We know,

Frequency,

$f = \dfrac{c}{\lambda }$
where,

$c =$ velocity of light

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

$\lambda =$ wavelenght

$\therefore f= \dfrac{ 3 \times 10^{8}}{5000 \times 10^{-10} }$

$f= 6 \times 10^{14}\ Hz$

The displacement of a particle varies according to the relation

$x=4(cos \pi t+ sin \pi t)$ . The amplitude of the particle is

0%
$-4 m$
0%
$4 m$
0%
$4\sqrt{2} m$
0%
$8m$

Explanation

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

$= 4[sin(\dfrac{\pi}{2}-\pi t)+ sin \pi t ]$

$= 8[ sin\dfrac{\pi}{4}. cos(-\dfrac{\pi}{4}+\pi t)]$ since

$[cos(-\theta)= cos\theta]$

$=\dfrac{8}{\sqrt{2}}. cos[\pi t- \dfrac{\pi}{4}] \\ = 4\sqrt{2} cos[\pi t- \dfrac{\pi}{4}]$

Comparing it with standard equation:

$x = A cos(\omega t- kx) \Rightarrow A=4\sqrt{2}$

A particle executing simple harmonic motion along y-axis has its motion described by the equation

$y = A\sin(\omega t) + B$ . The amplitude of the simple harmonic motion is:

0%
$A$
0%
$B$
0%
$A+B$
0%
$\sqrt{A+B}$

Explanation

The equation

$y = A sin (wt) + B$ suggests that the particle is executing Simple Harmonic Motion about

$B$ which is the mean position and the amplitude of oscillation of the SHM (about

$B$ ) is equal to

$A$ .

Consider a sinusoidal travelling wave shown in figure. The wave velocity is

$+ 40 cm/s$ . Find the frequency.

0%
$2 Hz$
0%
$6 Hz$
0%
$8 Hz$
0%
$10 Hz$

Explanation

Clearly from figure, the wavelength is equal to the distance between two in-phase points.
Let us consider the points where it cuts the x-axis.

The in-phase points cut x-axis at

$2.5\ cm$ and

$6.5\ cm$ .

Thus the wavelength is

$4cm$
We have

$v=\lambda\nu$

$\implies \nu=\dfrac{v}{\lambda}=\dfrac{40cm/s}{4cm}=10Hz$

Sound waves of wavelength

$\lambda$ travelling with velocity

$v$ in a medium enter into another medium in which their velocity is

$4v$ . The wavelength in

$2^{nd}$ medium is :

0%
$4\lambda$
0%
$\lambda$
0%
$\lambda/4$
0%
$16\lambda$

Explanation

From

$v=n\lambda$ we find

$\lambda\propto v$ because frequency n is constant .

Therefore ,

new wavelength =

$4\lambda$

A string of length

$20\ cm$ and linear mass density

$0.4\ g/cm$ is fixed at both ends and is kept under a tension of

$16 \ N$ . A wave pulse is produced at

$t = 0$ near an end as shown in figure which travels towards the other end. The string have the shape shown in the figure again in

$2 \times 10^{-x} sec$ . Find

$x$ .

0%
$1$
0%
$2$
0%
$8$
0%
$3$

Explanation

Given :

$T =16$ N

$L =0.2$ m

$\mu =0.4$ g/cm

$= 0.04$ kg/m

Velocity of wave

$v = \sqrt{\dfrac{T}{\mu}} = \sqrt{\dfrac{16}{0.04}} =20$ m/s

Time after which the string has the same shape

$t = \dfrac{2L}{v}$

$\therefore$

$t = \dfrac{2(0.2)}{20}= 2\times 10^{-2}$ s

$\implies x = 2$

State whether the given statement is True or False :

Consider a sinusoidal travelling wave shown in figure. The wave velocity is

$+ 40 cm/s$ . The phase difference between points

$2.5\ cm$ apart will be

$\dfrac{5\pi}{4}$ rad.

0%
True
0%
False

Explanation

Clearly from figure, the wavelength is equal to the distance between two in-phase points.
Let us consider the points where it cuts the x-axis.

The in-phase points cut x-axis at 2.5cm and 6.5cm.

Thus the wavelength is

$4cm$

These points have a phase difference of

$2\pi$ .

Thus phase difference between points

$2.5cm$ apart=

$\dfrac{2.5}{4}\times 2\pi$

$rad=\dfrac{5\pi}{4}rad$

Two waves are travelling in the same direction along a stretched string. The waves are

$90^0$ out of phase. Each wave has an amplitude of 4.0 cm. The amplitude of the resultant wave is

$5.66 \times 10^{-x} m$ . Find x.

0%
1
0%
2
0%
8
0%
4

Explanation

Let A be the amplitude of each wave. Thus the waves are

$y_1=Acos(\omega t-kx)$

and

$y_2=Asin(\omega t-kx)$

Thus the net wave is

$y= y_1 + y_2 = Acos(\omega t-kx)+Asin(\omega t-kx)$

$=\sqrt{2}Asin(\omega t-kx+\dfrac{\pi}{4})$

Thus the new amplitude is

$\sqrt{2}A=5.66cm = 5.66\times 10^{-2} m$

Thus, x=2.

The equation of a wave is given by

$Y\, =\, 5\, sin\, 10 \pi\, (t\, -\, 0.01x)$ along the x-axis. (All the quantities are expressed in SI units}. The phase difference the points separated by a distance of 10 m along x-axis is

0%
$\displaystyle \frac{\pi}{2}$
0%
$\pi$
0%
$2 \pi$
0%
$\displaystyle \frac{\pi}{4}$

Explanation

For the given wave,

$k=0.1\pi=\dfrac{2\pi}{\lambda}$

$\implies \lambda=20m$

Thus phase difference between two points separated by 10m is

$\dfrac{2\pi}{\lambda}(x_2-x_1)$

$=\dfrac{2\pi}{\lambda}\times 10m=\pi$

Hence correct answer is option B.

The period of a particle in SHM is

$8 s$ . At

$t=0$ it is in its equilibrium position. Find the ratio of the distance traveled in the first

$2 s$ and the next

$2 s$ .

0%
$3$
0%
$5$
0%
$2$
0%
$1$

Three coherent waves having amplitudes 12 mm, 6 mm and 4 mm arrive at a given point with successive phase difference of

$\pi/2$ . Then the amplitude of the resultant wave is

0%
7 mm
0%
10 mm
0%
5 mm
0%
4.8 mm

Explanation

Since, it is given that the waves arrive at the same point with a phase difference of

$\pi/2$

Therefore, the waves having an amplitude of 12mm and 4 mm are completely out of phase, and then their resultant amplitude is 12 - 4 = 8mm

Now, the 8mm and 6mm waves are

$90^o$ out of phase so, their resultant amplitude would be given by Pythagorus theorem

$A^2 = 6^2 + 8^2$

Therefore, A = 10 mm

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

$Y\, = \,A\, sin\,(kx\,-\,\omega\,t+\,30^{\circ})$ . It is reflected from a heavy string tied to end of the light string at x = 0 . If 64% of the incident energy is reflected then the equation of the reflected wave is

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

Explanation

There are three things we need to take into account:

Energy transfer
Change in velocity
Change in phase

We know that power delivered is proportional to

$A^{2}$
Hence if power(energy) reduces to 64%. We get that Amplitude must reduce to 80% or 0.8A.
Now the reflected wave is moving in the opposite direction. (velocity is negative now).
Also because of the hard soft boundary reflection (there is a phase lag of

$180^{\circ}$
Hence the new equation becomes:

$y = 0.8A sin(kx + \omega t + 30^{\circ} + 180^{\circ})$
Hence option B.

A source oscillates with a frequency 25 Hz and the wave propagates with 300 m/s. Two points A and B are located at distances 10 m and 16 m away from the source. The phase difference between A and B is

0%
$\displaystyle \frac{\pi}{4}$
0%
$\displaystyle \frac{\pi}{2}$
0%
$\pi$
0%
$2 \pi$

Explanation

Wavelength of the wave=

$\lambda=\dfrac{v}{\nu}=\dfrac{300}{25}=12m$

Distance between the two points=

$16m-10m=6m=\dfrac{\lambda}{2}$

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

Three one dimensional mechanical waves in an elastic medium is given as and

$y_1\,=\,3A \,sin\,(\omega\,t\,-\,kx)$

$y_2\,=\,A \,sin\,(\omega\,t\,-\,kx\,+\,\pi)$

$y_3\,=\,2A \,sin\,(\omega\,t\,+\,kx)$
are superimposed with each other. The maximum displacement amplitude of the medium particle would be

0%
4A
0%
3A
0%
2A
0%
A

Vibrations of period 0.25 s propagate along a straight line at a velocity of 48 cm/s. One second after the emergence of vibrations at the initial point, displacement of the point, 47 cm from it is found to be 3 cm. Then,

0%
amplitude of vibrations is 6 cm.
0%
amplitude of vibrations is $3 \sqrt{2} cm.$
0%
amplitude of vibrations is 3 cm.
0%
None of the above

Explanation

The wavelength of the wave can be found by using

$\dfrac{\lambda}{T}=v$

$\implies \lambda=vT=48cm/s\times 0.25s=12cm$

Four full wavelengths complete at a distance of 48cm.

Thus a point 47cm lag by a phase difference of

$\dfrac{2\pi}{\lambda}(48cm-47cm)=\dfrac{\pi}{6}$

Let the amplitude of vibrations be

$A$ .

Thus the displacement at the given point=

$Asin(\dfrac{\pi}{6})=\dfrac{A}{2}=3cm$

$\implies A=6cm$

Thus correct answer is option A.

A body is vibrating 7200 times in one minute. If the velocity of sound is 360 m/s, find (i) frequency of the vibration in Hz, (ii) the wavelength of the sound produced.

0%
$120\ Hz, 3\ m$
0%
$140\ Hz, 3\ m$
0%
$120\ Hz, 4\ m$
0%
$140\ Hz, 4\ m$

Explanation

(i) Frequency of vibration is defined as number of vibration per second ,

so , frequency ,

$f=$ number of vibrations/time in second

$=7200/60=120Hz$ ,

(ii) Now given ,

$v=360m/s$ and

$f=120Hz$

by ,

$v=f\lambda$ ,

$\lambda=v/f=360/120=3m$

A .......... is any disturbance that transmits energy without carrying any material with it.

0%
Wave
0%
Temperature
0%
Vibration
0%
Echolocation

A particle moves on the X-axis is according to the equation

$x = A+B\space \sin\omega t$ . The motion is simple harmonic with amplitude:

0%
$A$
0%
$B$
0%
$A+B$
0%
$\sqrt{A^2 + B^2}$

Explanation

The equation

$y = B sin (wt) + A$ suggests that the particle is executing Simple Harmonic Motion about

$A$ which is the mean position and the amplitude of oscillation of the SHM (about

$A$ ) is equal to

$B$ .

Sound travels with a speed of about 330m/s. What is the wavelength of sound whose frequency is 660 Hz?

0%
0.5 m
0%
1 m
0%
2 m
0%
1.5 m

Explanation

We have ,

$v=f\lambda$ ,

given

$v=330m/s , f=660Hz$ ,

therefore ,

$\lambda=v/f=330/660=0.5m$

Which of the following parameters of a wave undergoes a change when wave is reflected from a boundary ?

0%
$Intensity$
0%
$Phase$
0%
$Speed$
0%
$Frequency$

CBSE Questions for Class 11 Engineering Physics Waves Quiz 5 - MCQExams.com

0:0:1

Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Waves Quiz 5 - MCQExams.com

0:0:1

Practice Class 11 Engineering Physics Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.