Waves — Page 3 | JEE Physics

Q21

The displacement

y

of a wave travelling in the

x

-direction is given by

y = {10^{ - 4}}\,\sin \left( {600t - 2x + {\pi \over 3}} \right)\,\,metres

where

x

is expressed in metres and

t

in seconds. The speed of the wave - motion, in

m{s^{ - 1}}

$, is

A

300

B

600

C

1200

D

200

Correct Answer

Option A

Solution

y = {10^{ - 4}}\sin \left( {600t - 2x + {\pi \over 3}} \right)

But

y = A\sin \left( {\omega t - kx + \phi } \right)

On comparing we get

\omega = 600;\,k = 2

v = {\omega \over k} = {{600} \over 2} = 300\,m{s^{ - 1}}

Q22

A whistle producing sound waves of frequencies

9500

Hz

and above is approaching a stationary person with speed

v

m{s^{ - 1}}.

The velocity of sound in air is

300\,m{s^{ - 1}}.

If the person can hear frequencies upto a maximum of

10,000

HZ,

the maximum value of

v

upto which he can hear whistle is

A

15\sqrt 2 \,\,m{s^{ - 1}}

B

{{15} \over {\sqrt 2 }}\,m{s^{ - 1}}

C

15\,\,m{s^{ - 1}}

D

30\,\,m{s^{ - 1}}

Correct Answer

Option C

Solution

v' = v\left[ {{v \over {v - {v_s}}}} \right] \Rightarrow 10000

= 9500\left[ {{{300} \over {300 - v}}} \right]

\Rightarrow 300 - v = 300 \times 0.95 \Rightarrow v

= 300 - 285 = 15\,m{s^{ - 1}}

Q23

A string is stretched between fixed points separated by

75.0

cm.

It is observed to have resonant frequencies of

420

Hz

and

315

Hz

. There are no other resonant frequencies between these two. Then, the lowest resonant frequency for this string is

A

105

Hz

B

1.05

Hz

C

1050

Hz

D

10.5

Hz

Correct Answer

Option A

Solution

Given

{{nv} \over {2\ell }} = 315

and

\left( {n + 1} \right){v \over {2\ell }} = 420

\Rightarrow {{n + 1} \over n} = {{420} \over {315}} \Rightarrow n = 3

Hence

3 \times {v \over {2\ell }} = 315 \Rightarrow {v \over {2\ell }} = 105Hz

Lowest resonant frequency is when

n=1

Therefore lowest resonant frequency

= 105\,Hz.

Q24

Three sound waves of equal amplitudes have frequencies

\left( {v - 1} \right),\,v,\,\left( {v + 1} \right).

They superpose to give beats. The number of beats produced per second will be :

A

3

B

2

C

1

D

4

Correct Answer

Option B

Solution

Maximum number of beats

= \left( {v + 1} \right) - \left( {v - 1} \right) = 2

Q25

A motor cycle starts from rest and accelerates along a straight path at

2m/{s^2}.

At the starting point of the motor cycle there is a stationary electric siren. How far has the motor cycle gone when the driver hears the frequency of the siren at

94\%

of its value when the motor cycle was at rest? (Speed of sound

= 330\,m{s^{ - 1}}

)

A

98

m

B

147

m

C

196\,m

D

49

m

Correct Answer

Option A

Solution

v_m^2 - {u^2} = 2as \Rightarrow v_m^2 = 2 \times 2 \times s

$\therefore$

{v_m} = 2\sqrt s

According to Doppler's effect

0.94v = v\left[ {{{330 - 2\sqrt s } \over {330}}} \right] \Rightarrow s = 98.01\,m

Q26

The equation of a wave on a string of linear mass density

0.04\,\,kg\,{m^{ - 1}}

is given by

y = 0.02\left( m \right)\,\sin \left[ {2\pi \left( {{t \over {0.04\left( s \right)}} - {x \over {0.50\left( m \right)}}} \right)} \right].

$ The tension in the string is

A

4.0N

B

12.5

N

C

0.5

N

D

6.25

N

Correct Answer

Option D

Solution

y = 0.02\left( m \right)\sin \left[ {2\pi \left( {{t \over {0.04\left( s \right)}}} \right) - {x \over {0.50\left( m \right)}}} \right]

But

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

$\therefore$

\omega = {{2\pi } \over {0.04}} \Rightarrow v = {1 \over {0.04}} = 25\,Hz

k = {{2\pi } \over {0.50}} \Rightarrow \lambda = 0.5m

$\therefore$ velocity,

v = v\lambda = 25 \times 0.5\,m/s = 12.5\,m/s

Velocity on a string is given by

v = \sqrt {{T \over \mu }}

$\therefore$

T = {v^2} \times \mu = {\left( {12.5} \right)^2} \times 0.04 = 6.25\,N

Q27

The transverse displacement

y(x, t)

of a wave on a string is given by

y\left( {x,t} \right) = {e^{ - \left( {a{x^2} + b{t^2} + 2\sqrt {ab} \,xt} \right)}}.

This represents

a:

A wave moving in

-x

direction with speed

\sqrt {{b \over a}}

B standing wave of frequency

\sqrt b

C standing wave of frequency

{1 \over {\sqrt b }}

D wave moving in

+x

direction speed

\sqrt {{a \over b}}

Correct Answer

Option A

Solution

Given wave equation is

y\left( {x,t} \right){ = _e}\left( { - a{x^2} + b{t^2} + 2\sqrt {ab} \,xt} \right)

= {e^{ - \left[ {{{\left( {\sqrt {ax} } \right)}^2} + {{\left( {\sqrt {bt} } \right)}^2} + 2\sqrt a x.\sqrt b t} \right]}}

= {e^{ - {{\left( {\sqrt a x + \sqrt b t} \right)}^2}}}

= {e^{ - {{\left( {x + \sqrt {{b \over a}} t} \right)}^2}}}

It is a function of type

y = f\left( {x + vt} \right)

$\Rightarrow$ Speed of wave

= \sqrt {{b \over a}}

Q28

A sonometer wire of length

1.5

m

is made of steel. The tension in it produces an elastic strain of

1\%

. What is the fundamental frequency of steel if density and elasticity of steel are

7.7 \times {10^3}\,kg/{m^3}

and

2.2 \times {10^{11}}\,N/{m^2}

respectively ?

A

188.5

Hz

B

178.2

Hz

C

200.5

Hz

D

770

Hz

Correct Answer

Option B

Solution

Fundamental frequency,

f = {v \over {2\ell }} = {1 \over {2\ell }}\sqrt {{T \over \mu }} = {1 \over {2\ell }}\sqrt {{T \over {A\rho }}}

\left[ {\,\,} \right.

as

v = \sqrt {{T \over \mu }}

\,\,\,\,\,\,

and

\,\,\,\,\,\,

\left. {\mu = {m \over \ell }\,\,} \right]

Also,

Y = {{T\ell } \over {A\Delta \ell }} \Rightarrow {T \over A} = {{Y\Delta \ell } \over \ell }

\Rightarrow f = {1 \over {2\ell }}\sqrt {{{\gamma \Delta \ell } \over {\ell \rho }}} ....\left( i \right)

Putting the value of

\ell ,{{\Delta \ell } \over \ell },\rho

\,\,\,\,\,\,

and

\,\,\,\,\,\,

$\gamma$ in

e{q^n}.\left( i \right)

we get,

f = \sqrt {{2 \over 7}} \times {{{{10}^3}} \over 3}

\,\,\,\,\,\,

or,

\,\,\,\,\,\,

f \approx 178.2\,Hz

Q29

A train is moving on a straight track with speed

20\,m{s^{ - 1}}.

It is blowing its whistle at the frequency of

1000

Hz

. The percentage change in the frequency heard by a person standing near the track as the train passes him is (speed of sound

= 320\,m{s^{ - 1}}

) close to :

A

18\%

B

24\%

C

6\%

D

12\%

Correct Answer

Option D

Solution

{f_1} = f\left[ {{v \over {v - {v_s}}}} \right] = f \times {{320} \over {300}}Hz

{f_2} = f\left[ {{v \over {v + {v_s}}}} \right] = f \times {{320} \over {340}}Hz

\left( {{{{f_2}} \over {{f_1}}} - 1} \right) \times 100 = \left( {{{300} \over {340}} - 1} \right) \times 100 = 12\%

Q30

A pipe open at both ends has a fundamental frequency

f

in air. The pipe is dipped vertically in water so that half of it is in water. The fundamental frequency of the air column is now :

A

2f

B

f

C

{f \over 2}

D

{3f \over 4}

Correct Answer

Option B

Solution

The fundamental frequency in case

(a)

is

f = {v \over {2\ell }}

The fundamental frequency in case

(b)

is

f'{v \over {4\left( {\ell /2} \right)}} = {u \over {2\ell }} = f