Waves — Page 6 | NEET Physics

Q51

The phase difference between two waves. represented by y 1 = 10

-

6 sin[100t + (x/50) + 0.5] m y 2 = 10

-

6 cos[100t + (x/50)] m, where x is expressed in metres and t is exressed in secondss, is approximately.

A 1.07 radians

B 2.07 radians

C 0.5 radians

D 1.5 radians

Correct Answer

Option A

Solution

y 1 = 10 –6 sin[100t + (x/50) + 0.5] y 2 = 10 –6 cos[100t + (x/50)] [using cosx = sin(x + $\pi$ /2)] = 10 –6 sin[100t + (x/50) + $\pi$ /2] = 10 –6 sin[100t + (x/50) + 1.57] The phase difference = 1.57 – 0.5 = 1.07 [or using sinx = cos( $\pi$ /2 – x).

We get the same result.]

Q52

A car is moving towards a high cliff. The driver sounds a horn of frequency

f

. The reflected sound heard by the driver has frequency

2f

. If v is the velocity of sound, then the velocity of the car, in the same velocity units, will be

A v/

\sqrt 2

B v/3

C v/4

D v/2

Correct Answer

Option B

Solution

Let f' be the frequency of sound heard by cliff. $\therefore$

f = {{vf} \over {v - {v_c}}}

...(1) Now, for the reflected wave, cliff acts as a source, $\therefore$

2f = {{f'\left( {v + {v_c}} \right)} \over v}

...(2)

2f = {{f\left( {v + {v_c}} \right)} \over {v - {v_c}}}

\Rightarrow 2v - 2{v_c} = v + {v_c}

\Rightarrow {v \over 3} = {v_c}

Q53

An observer moves towards a stationary source of sound with a speed 1/5 th of the speed of sound. The wavelength and frequency of the source emitted are

\lambda

and

f

respectively. The apparent frequency and wavelength recorded by the observer are respectively

A 1.2

f

, 1.2

\lambda

B 1.2

f

,

\lambda

C

f

, 1.2

\lambda

D 0.8

f

, 0.8

\lambda

Correct Answer

Option B

Solution

Apparent frequency,

f' = {{v + {v_0}} \over v}f

= {{v + (1/5)v} \over v}f = 1.2f

Wavelength does not change by motion of observer.

Q54

A wave travelling in positive X-direction with a

=

0.2 ms

-

2 , velocity = 360 ms

-

1 and

\lambda

=

60 m, then correct expression for the wave is

A

y = 0.2\sin \left[ {2\pi \left( {6t + {x \over {60}}} \right)} \right]

B

y = 0.2\sin \left[ {\pi \left( {6t + {x \over {60}}} \right)} \right]

C

y = 0.2\sin \left[ {2\pi \left( {6t - {x \over {60}}} \right)} \right]

D

y = 0.2\sin \left[ {\pi \left( {6t - {x \over {60}}} \right)} \right]

Correct Answer

Option C

Solution

The equation of progressive wave travelling in positive x-direction is given by

y = a\sin {{2\pi } \over \lambda }\left( {vt - x} \right)

Here a = 0.2 m, v = 360 m/sec, $\lambda$ = 60 m

\therefore y = 0.2\sin {{2\pi } \over {60}}\left( {360t - x} \right)

= 0.2sin\left[ {2\pi \left( {6t - {x \over {60}}} \right)} \right]

Q55

A whistle revolves in a circle with angular speed

\omega

= 20 rad/s using a string of length 50 cm. If the frequency of sound from the whistle is 385 Hz, then what is the minimum frequency heard by an observer which is far away from the centre (velocity of sound

=

340 m/s)

A 385 Hz

B 374 Hz

C 394 Hz

D 333 Hz.

Correct Answer

Option B

Solution

The whistle is revolving in a circle of radius 50 cm.

So the source (whistle) is moving and the observer is fixed.

The minimum frequency will be heard by the observer when the linear velocity of the whistle (source) will be in a direction as shown in the figure, i.e. when the source is receding.

The apparent frequency heard by the observer is then given by

\upsilon ' = \upsilon \left( {{v \over {V + v}}} \right)

where V and v are the velocities of sound and source respectively and

\upsilon

is the actual frequency. Now

v = r\omega = 0.5 \times 20 = 10\,m/s

V = 340 m/s,

\upsilon

= 385 Hz. $\therefore$

\upsilon ' = 385 \times {{340} \over {340 + 10}} = 374

Hz

Q56

Two waves having equation x 1 =

a

sin(

\omega

t

-

kx +

\phi

1 ), x 2 = asin(

\omega

t

-

kx +

\phi

2 ). If in the resultant wave the frequency and amplitude remain equal to amplitude of superimposing waves, the phase difference between them is

A

{\pi \over 6}

B

{{2\pi } \over 3}

C

{\pi \over 4}

D

{\pi \over 3}

Correct Answer

Option B

Solution

Resultant amplitude = 2a (1 + cos $\phi$ ) = a $\therefore$

\left( {1 + \cos \phi } \right) = 1/2;

cos\phi = - {1 \over 2};\phi = {{2\pi } \over 3}

Q57

If the tension and diameter of a sonometer wire of fundamental frequency n is doubled and density is halved then its fundamental frequency will become

A

{\pi \over 4}

B

\sqrt 2 n

C n

D

{n \over {\sqrt 2 }}

Correct Answer

Option C

Solution

n = {1 \over {2l}}\sqrt {{T \over {\pi {r^2}\rho }}}

\rho _1^{'} = {\rho \over 2};T' = 2T

D' = 2D or r' = 2r

n' = {1 \over {2l}}\sqrt {{{2T} \over {\pi {{\left( {2r} \right)}^2}{\rho \over 2}}}}

After solving,

n' = {1 \over {2l}}\sqrt {{{2T} \over {\pi {r^2}l}}} = n

Q58

The equation of a wave is represented by y

=

10

-

4 sin(100t

-

{x \over {10}}

) m. then the velocity of wave will be

A 100 m/s

B 4 m/s

C 1000 m/s

D 10 m/s

Correct Answer

Option C

Solution

Comparing the given equation with general equation,

y = a\sin 2\pi \left( {{t \over T} - {x \over \lambda }} \right)

, we get

T = {{2\pi } \over {100}}

and

\lambda = 20\pi

\therefore v = \upsilon \lambda = {{100} \over {2\pi }} \times 20\pi = 1000\,m/s

Q59

A string is cut into three parts, having fundamental frequencies n 1 , n 2 , n 3 respectively. Then original fundamental frequency n related by the expression as

A

{1 \over n} = {1 \over {{n_1}}} + {1 \over {{n_2}}} + {1 \over {{n_3}}}

B

n = {n_1} \times {n_2} \times {n_3}

C n

=

n 1 + n 2 + n 3

D

n = {{{n_1} + {n_2} + {n_3}} \over 3}

Correct Answer

Option A

Solution

As

n \propto \left( {1/l} \right)

l = {l_1} + {l_2} + {l_3}

\therefore {1 \over n} = {1 \over {{n_1}}} + {1 \over {{n_2}}} + {1 \over {{n_3}}}

Q60

The equations of two waves acting in perpendicular directions are given as x =

a

cos(

\omega

t +

\delta

) and y =

a

cos(

\omega

t +

\alpha

), where

\delta

=

\alpha

+

{\pi \over 2}

, the resultant wave represents

A a parabola

B a circle

C an ellipse

D a straight line

Correct Answer

Option B

Solution

Given : x = acos( $\omega$ t + $\delta$ ) and y = acos( $\omega$ t + $\alpha$ ) ...(i) where,

\delta = \alpha + \pi /2

\therefore {\rm{x = acos(}}\omega {\rm{t + }}\alpha {\rm{ + }}\pi {\rm{/2)}}

= –asin( $\omega$ t + $\alpha$ ) ...(ii) Given the two waves are acting in perpendicular direction with the same frequency and phase difference $\pi$ /2.

From equations (i) and (ii), x 2 + y 2 = a 2 which represents the equation of a circle.