Waves — Page 9 | JEE Physics

Q81

Assume that the displacement(s) of air is proportional to the pressure difference (

\Delta

p) created by a sound wave. Displacement (s) further depends on the speed of sound (v), density of air (

\rho

) and the frequency (f). If

\Delta

p ~ 10 Pa, v ~ 300 m/s,

\rho

~ 1 kg/m3 and f ~ 1000 Hz, then s will be of the order of (take the multiplicative constant to be 1) :

A 1 mm

B

{3 \over {100}}

mm

C 10 mm

D

{1 \over {10}}

mm

Correct Answer

Option B

Solution

Given, S $\propto$

\Delta

p and Proportionally constant = 1 We know,

\Delta

p = S $\beta$ k = $\rho$ v2 $\times$

{\omega \over v} \times S

=

{\rho v\omega S}

$\therefore$ S =

{{\Delta p} \over {\rho v\omega }}

=

{{\Delta p} \over {\rho v2\pi f}}

=

{{\Delta p} \over {\rho vf}}

[As Proportionally constant = 1 so assume 2 $\pi$ = 1]

= {{10} \over {1 \times 300 \times 1000}}

= {1 \over {30}}mm

\approx {3 \over {100}}mm

Q82

Two sitar strings, A and B, playing the note 'Dha' are slightly out of tune and produce beats of frequency 5 Hz. The tension of the string B s slightly increased and the beat frequency is found to decrease by 3Hz. If the frequency of A is 425 Hz, the original frequency of B is :

A 430 Hz

B 420 Hz

C 428 Hz

D 422 Hz

Correct Answer

Option B

Solution

Frequency of B, fB = 425 $\pm$ 5 = 420 or 430 Hz As tension of string B is increased So, frequency of B, fB should also increase [as f $\propto$

\sqrt T

] If initially fB = 430 Hz then when fB increases by increasing the tension then fB $-$ fA increases that means beat frequency increase.

So, fB can't be 430 Hz When fB = 420 then when fB increases fA $-$ fB decreases means beat frequency decreases.

So, correct fB = 420 Hz.

Q83

An observer moves towards a stationary source of sound with a velocity equal to one-fifth of the velocity of sound. The percentage change in the frequency will be :

A 20%

B 10%

C 5%

D 0%

Correct Answer

Option A

Solution

f' = {f_0}\left[ {{{v - {v_0}} \over {v - {v_s}}}} \right]

\Rightarrow f' = {f_0}\left[ {{{v + {v \over 5}} \over v}} \right]

\Rightarrow f' = {{6{f_0}} \over 5}

$\Rightarrow$ % change = 20

Q84

Two engines pass each other moving in opposite directions with uniform speed of 30 m/s. One of them is blowing a whistle of frequency 540 Hz. Calculate the frequency heard by driver of second engine before they pass each other. Speed of sound is 330 m/sec :

A 450 Hz

B 540 Hz

C 648 Hz

D 270 Hz

Correct Answer

Option C

Solution

Frequency heard by the driver of second engine, F' =

\left( {{{v + {v_0}} \over {v - {v_S}}}} \right)

f given v0 = vs = 30 m/s $\therefore$ f' =

\left( {{{330 + 30} \over {330 - 30}}} \right) \times 540

= 648 Hz

Q85

A granite rod of 60 cm length is clamped at its middle point and is set into longitudinal vibrations. The density of granite is 2.7

\times

103 kg/m3 and its Young’s modulus is 9.27

\times

1010 Pa. What will be the fundamental frequency of the longitudinal vibrations ?

A 7.5 kHz

B 5 kHz

C 2.5 kHz

D 10 kHz

Correct Answer

Option B

Solution

As rod length = 60 cm

\therefore\,\,\,

{\lambda \over 2}

= 60 $\Rightarrow$

\,\,\,

$\lambda$ = 120 cm = 1.2 m In solid, velocity of wave, V =

\sqrt {{Y \over \rho }}

=

\sqrt {{{9.27 \times {{10}^{10}}} \over {2.7 \times {{10}^3}}}}

= 5.85 $\times$ 103 m/sec. As we know, v = f $\lambda$

\therefore\,\,\,

f =

{v \over \lambda }

=

{{5.85 \times {{10}^3}} \over {1.2}}

= 4.88 $\times$ 103 Hz

\simeq

5 kHz

Q86

A sound absorber attenuates the sound level by

20

dB

. The intensity decreases by a factor of

A

100

B

1000

C

10000

D

10

Correct Answer

Option A

Solution

We have,

{L_1} = 10\log \left( {{{{{\rm I}_1}} \over {{{\rm I}_0}}}} \right);

{L_2} = 10\,\log \left( {{{{{\rm I}_2}} \over {{{\rm I}_0}}}} \right)

$\therefore$

\,\,{L_1} - {L_2} = 10\,\log \left( {{{{{\rm I}_1}} \over {{{\rm I}_0}}}} \right) - 10\,\log \left( {{{{{\rm I}_2}} \over {{{\rm I}_0}}}} \right)

or,

\Delta L = 10\,\log \left( {{{{{\rm I}_1}} \over {{{\rm I}_0}}} \times {{{{\rm I}_0}} \over {{{\rm I}_2}}}} \right)

or,

\Delta L = 10\,\log \left( {{{{{\rm I}_1}} \over {{{\rm I}_2}}}} \right)

or,

20 = 10\log \left( {{{{{\rm I}_1}} \over {{{\rm I}_2}}}} \right)

or,

2 = \log \left( {{{{{\rm I}_1}} \over {{{\rm I}_2}}}} \right)

or,

{{{{\rm I}_1}} \over {{{\rm I}_2}}} = {10^2}

or,

{{\rm I}_2} = {{{{\rm I}_1}} \over {100}}.

$\Rightarrow$ Intensity decreases by a factor

100.

Q87

A travelling harmonic wave is represented by the equation y(x,t) = 10–3sin (50t + 2x), where, x and y are in mater and t is in seconds. Which of the following is a correct statement about the wave ?

A The wave is propagating along the positive x-axis with speed 100 ms–1

B The wave is propagating along the positive x-axis with speed 25 ms–1

C The wave is propagating along the negative x-axis with speed 25 ms–1

D The wave is propagating along the negative x-axis with speed 100 ms–1

Correct Answer

Option C

Solution

y = a sin( $\omega$ t + kx) $\Rightarrow$ wave is moving along $-$ ve x-axis with speed v =

{\omega \over K}

$\Rightarrow$ v =

{{50} \over 2}

= 25m/sec

Q88

A standing wave is formed by the superposition of two waves travelling in opposite directions. The transverse displacement is given by y(x, t) = 0.5 sin

\left( {{{5\pi } \over 4}x} \right)\,

cos(200

\pi

t). What is the speed of the travelling wave moving in the positive x direction ? (x and t are in meter and second, respectively.)

A 160 m/s

B 90 m/s

C 180 m/s

D 120 m/s

Correct Answer

Option A

Solution

Standard equation of standing wave, y(x, t) = 2a sin kx cos $\omega$ t Given, y(x, t) = 0.5 sin

\left( {{{5\pi } \over 4}x} \right)

cos (200 $\pi$ t). So, k =

{{5\pi } \over 4}

and $\omega$ = 200 $\pi$

\therefore\,\,\,

Speed of travelling wave =

{\omega \over k}

=

{{200\pi } \over {{{5\pi } \over 4}}}

= 160 m/s.

Q89

A source of sound S is moving with a velocity of 50 m/s towards a stationary observer. The observer measures the frequency of the source as 1000 Hz. What will be the apparent frequency of the source when it is moving away from the observer after crossing him? (Take velocity of sound in air is 350 m/s)

A 750 Hz

B 857 Hz

C 807 Hz

D 1143 Hz

Correct Answer

Option A

Solution

{f_a} = {V \over {V - {V_s}}}{f_o} = 1000\,Hz

f_a^{'} = {V \over {V + {V_s}}}{f_o}

{{f_a^{'}} \over {{f_a}}} = {{V - {V_s}} \over {V + {V_s}}} = {{350 - 50} \over {350 + 50}} = {{300} \over {400}} = {3 \over 4}

f_a^{'} = {3 \over 4} \times 1000 = 750\,Hz

Q90

A sound wave of frequency 245 Hz travels with the speed of 300 ms

-

1 along the positive x-axis. Each point of the wave moves to and from through a total distance of 6 cm. What will be the mathematical expression of this travelling wave?

A Y(x, t) = 0.03 [ sin 5.1x

-

(0.2

\times

103)t ]

B Y(x, t) = 0.03 [ sin 5.1x

-

(1.5

\times

103)t ]

C Y(x, t) = 0.06 [ sin 5.1x

-

(1.5

\times

103)t ]

D Y(x, t) = 0.06 [ sin 0.8x

-

(0.5

\times

103)t ]

Correct Answer

Option B

Solution

Y = A\sin (kx - \omega t)

A = {6 \over 2}

= 3cm = 0.03 m

\omega = 2\pi f = 2\pi \times 245

\omega = 1.5 \times {10^3}

k = {\omega \over v} = {{1.5 \times {{10}^3}} \over {300}}

k = 5.1

y = 0.03\sin (5.1x - (1.5 \times {10^3})t)