Waves — Page 4 | NEET Physics

Q31

The equation of a simple harmonic wave is given by y = 3 sin

{\pi \over 2}

(50t

-

x), where x and y are in metres and t is in seconds. The ratio of maximum particle velocity to the wave velocity is

A 2

\pi

B

{3 \over 2}\pi

C

3\pi

D

{2 \over 3}\pi

Correct Answer

Option B

Solution

y = 3\sin {\pi \over 2}\left( {50t - x} \right)

y = 3\sin \left( {25\pi t - {\pi \over 2}x} \right)

on comparing with the standard wave equation

y = a\sin \left( {\omega t - kx} \right)

Wave velocity

v = {\omega \over k} = {{25\pi } \over {\pi /2}} = 50\,m/\sec

The velocity of particle

{v_p} = {{\partial y} \over {\partial t}} = 75\pi \cos \left( {25\pi t - {\pi \over 2}x} \right)

{v_{p_{\max}}} = 75\pi

then

{{{v_{{p_{\max }}}}} \over v} = {{75\pi } \over {50}} = {{3\pi } \over 2}

Q32

When a string is divided into three segments of length

l

1 ,

l

2 and

l

3 the fundamental frequencies of these three segments are

{\upsilon _1},{\upsilon _2}

and

{\upsilon _3}

respectively. The original fundamental frequency (

v

) of the string is

A

\sqrt v = \sqrt {{v_1}} + \sqrt {{v_2}} + \sqrt {{v_3}}

B

v = {v_1} + {v_2} + {v_3}

C

{1 \over v} = {1 \over {{v_1}}} + {1 \over {{v_2}}} + {1 \over {{v_3}}}

D

{1 \over {\sqrt v }} = {1 \over {\sqrt {{v_1}} }} + {1 \over {\sqrt {{v_2}} }} + {1 \over {\sqrt {{v_3}} }}

Correct Answer

Option C

Solution

Let

l

be the length of the string. Fundamental frequency is given by

\upsilon = {1 \over {2l}}\sqrt {{T \over \mu }}

\Rightarrow v \propto {1 \over l}

( $\because$ T and $\mu$ are constants) Here,

{l_1} = {k \over {{\upsilon _1}}},{l_2} = {k \over {{\upsilon _2}}},{l_3} = {k \over {{\upsilon _3}}}

and

l = {k \over \upsilon }

But

l = l_1 + l_2 + l_3

{1 \over \upsilon } = {1 \over {{\upsilon _1}}} + {1 \over {{\upsilon _2}}} + {1 \over {{\upsilon _3}}}

Q33

Two sources of sound placed close to each other, are emitting progressive waves given by y 1 = 4sin600

\pi

t and y 2 = 5sin608

\pi

t An observer located near these two sources of sound will hear

A 4 beats per second with intensity ratio 25 : 16 between waxing and waning.

B 8 beats per second with intensity ratio 25 : 16 between waxing and waning.

C 8 beats per second with intensity ratio 81 : 1 between waxing and warning.

D 4 beats per second with intensity ratio 81 : 1 between waxing and waning.

Correct Answer

Option D

Solution

2\pi {f_1} = 600\pi

{f_1} = 300

...(i)

2\pi {f_2} = 608\pi

{f_2} = 304

...(ii)

\left| {{f_1} - {f_2}} \right| = 4\,beats

{{{I_{\max }}} \over {{I_{m.n}}}} = {{{{\left( {{A_1} + {A_2}} \right)}^2}} \over {{{\left( {{A_1} + {A_2}} \right)}^2}}} = {{{{\left( {5 + 4} \right)}^2}} \over {{{\left( {5 - 4} \right)}^2}}} = {{81} \over 1}

where A 1 , A 2 are amplitudes of given two sound wave.

Q34

Two identical piano wires, kept under the same tension T have a fundamental frequency of 600 Hz. The fractional increase in the tension of one of the wires which will lead to occurrence of 6 beats/s when both the wires oscillate together would be

A 0.01

B 0.02

C 0.03

D 0.04

Correct Answer

Option B

Solution

As

\upsilon = {1 \over {2L}}\sqrt {{T \over \mu }}

\therefore {{\Delta \upsilon } \over \upsilon } = {1 \over 2}{{\Delta T} \over T}

{{\Delta T} \over T} = 2{{\Delta \upsilon } \over \upsilon } = 2 \times {6 \over {600}} = 0.02

Q35

Two waves are represented by the equations y 1 =

a

sin(

\omega

t + kx + 0.57) m and y 2 = acos(

\omega

t + kx) m, where x is in meter and t

in

sec. The phase difference between them is

A 1.0 radian

B 1.25 radian

C 1.57 radian

D 0.57 radian

Correct Answer

Option A

Solution

Here, y 1 = a sin ( $\omega$ t + kx + 0.57) and y 2 = a cos ( $\omega$ t + kx)

= a\sin \left[ {{\pi \over 2} + \left( {\omega t + kx} \right)} \right]

Phase difference,

\Delta \phi = {\phi _2} - {\phi _1}

= {\pi \over 2} - 0.57

= {{3.14} \over 2} - 0.57

= 1.57 – 0.57 = 1 radian

Q36

Sound waves travel at 350 m/s through a warm air and at 3500 m/s through brass. The wavelength of a 700 Hz acoustic wave as it enters brass from warm air

A decrease by a factor 10

B increase by a factor 20

C increase by a factor 10

D decrease by a factor 20

Correct Answer

Option C

Solution

We have, v = n $\lambda$

\Rightarrow v \propto \lambda

(as n remains constant) Thus, as v increases 10 times, $\lambda$ also increases 10 times.

Q37

A transverse wave is represented by y = Asin(

\omega

t

-

kx). For what value of the wavelength is the wave velocity equal to the maximum particle velocity ?

A

\pi A/2

B

\pi A

C 2

\pi A

D

A

Correct Answer

Option C

Solution

y = A sin ( $\omega$ t–kx) Particle velocity,

{v_p} = {{dy} \over {dt}} = A\omega \cos \left( {\omega t - kx} \right)

\therefore {v_{p\,\max }} = A\omega

wave velocity =

{\omega \over k}

$\therefore$

A\omega = {\omega \over k}

i.e,

A = {1 \over k}

But

k = {{2\pi } \over \lambda }

\therefore \lambda = 2\pi A

Q38

A tuning fork of frequency 512 Hz makes 4 beats per second with the vibrating string of a piano. The beat frequency decreases to 2 beats per sec when the tension in the piano string is slightly increased. The frequency of the piano string before increasing the tension was

A 510 Hz

B 514 Hz

C 516 Hz

D 508 Hz

Correct Answer

Option D

Solution

Let the frequencies of tuning fork and piano string be

{\upsilon _1}

and

{\upsilon _2}

respectively. $\therefore$

{\upsilon _2} = {\upsilon _1} \pm 4 = 512\,Hz \pm 4

= 516 Hz or 508 Hz Increase in the tension of a piano string increases its frequency. If

{\upsilon _2}

= 516 Hz, further increase in

{\upsilon _2}

, resulted in an increase in the beat frequency. But this is not given in the question. If

{\upsilon _2}

= 508 Hz, further increase in

{\upsilon _2}

resulted in decrease in the beat frequency.

This is given in the question.

When the beat frequency decreases to 2 beats per second.

Therefore, the frequency of the piano string before increasing the tension was 508 Hz.

Q39

The driver of a car travelling with speed 30 m/s towards a hill sounds a horn of frequency 600 Hz. If the velocity of sound in air is 330 m/s, the frequency of reflected sound as heard by driver is

A 555.5 Hz

B 720 Hz

C 500 Hz

D 550 Hz

Correct Answer

Option B

Solution

Car is the source and the hill is observer. Frequency heard at the hill,

{\upsilon _1}

\therefore {\upsilon _1} = {{\upsilon \times v} \over {\left( {v - V} \right)}} = {{600 \times 330} \over {330 - 30}}

Now for reflection, the hill is the source and the driver the observer. $\therefore$

{\upsilon _2} = {\upsilon _1} \times {{\left( {330 + 30} \right)} \over {330}}

\Rightarrow {\upsilon _2} = {{600 \times 330} \over {300}} \times {{360} \over {330}} \Rightarrow {\upsilon _2} = 720

Hz.

Q40

Each of the two strings of length 51.6 cm and 49.1 cm are tensioned separately by 20 N force. Mass per unit length of both the strings is same and equal to 1 g/m. When both the strings vibrate simultaneously the number of beats is

A 7

B 8

C 3

D 5

Correct Answer

Option A

Solution

l_1

= 0.516 m,

l_2

= 0.491 m, T = 20 N. Mass per unit length, $\mu$ = 0.001 kg/m. Frequency,

\upsilon = {1 \over {2l}}\sqrt {{T \over \mu }}

{\upsilon _1} = {1 \over {2 \times 0.516}}\sqrt {{{20} \over {0.001}}}

{\upsilon _2} = {1 \over {2 \times 0.491}}\sqrt {{{20} \over {0.001}}}

$\therefore$ Number of beats =

{\upsilon _1} - {\upsilon _2} = 7

.