Waves — Page 10 | JEE Physics

Q91

A stationary observer receives sound from two identical tuning forks, one of which approaches and the other one recedes with the same speed (much less than the speed of sound). The observer hears 2 beats/sec. The oscillation frequency of each tuning fork is v0 = 1400 Hz and the velocity of sound in air is 350 m/s. The speed of each tuning fork is close to :

A 1 m/s

B

{1 \over 8}

m/s

C

{1 \over 4}

m/s

D

{1 \over 2}

m/s

Correct Answer

Option C

Solution

f1 =

\left( {{c \over {c - v}}} \right){f_0}

f1 =

\left( {{c \over {c + v}}} \right){f_0}

beat frequency = f1 – f2 =

c{f_0}\left( {{1 \over {c - v}} - {1 \over {c + v}}} \right)

=

c{f_0}\left( {{{2v} \over {{c^2} - {v^2}}}} \right)

As c $\gg$ v then

{{c^2} - {v^2}}

=

{{c^2}}

=

c{f_0}\left( {{{2v} \over {{c^2}}}} \right)

=

{f_0}\left( {{{2v} \over c}} \right)

$\therefore$

{f_0}\left( {{{2v} \over c}} \right)

= 2 $\Rightarrow$ v =

{{350} \over {1400}}

=

{1 \over 4}

m/s

Q92

A tuning fork vibrates with frequency

256

Hz

and gives one beat per second with the third normal mode of vibration of an open pipe. What is the length of the pipe ? (Speed of sound in air is

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

)

A

220

cm

B

190

cm

C

180

cm

D

200

cm

Correct Answer

Option D

Solution

The tuning fork vibrates with frequency 256 Hz and give one beat per second So, the organ pipe will have frequency (256 $\pm$ 1) Hr.

For open organ pipe, Frequency n =

{{N\upsilon } \over {2\ell }}

Here n = 255 Hz N = 3

\upsilon

= 340 m/s

\therefore\,\,\,\,

255 =

{{3 \times 340} \over {2 \times \ell }}

$\Rightarrow$

\,\,\,\,

\ell

=

{{3 \times 340} \over {2 \times 255}} = 2\,m

\therefore\,\,\,\,

\ell

= 2m or 200 cm

Q93

The equation of a transverse wave travelling along a string is

y(x, t)=4.0 \sin \left[20 \times 10^{-3} x+600 t\right] \mathrm{mm}

, where

x

is in mm and

t

is in second. The velocity of the wave is :

A

-60 \mathrm{~m} / \mathrm{s}

B

+60 \mathrm{~m} / \mathrm{s}

C

+30 \mathrm{~m} / \mathrm{s}

D

-30 \mathrm{~m} / \mathrm{s}

Correct Answer

Option D

Solution

Let's analyze the wave equation step by step. The given wave is:

y(x, t) = 4.0 \sin \left[20 \times 10^{-3} x + 600 t\right] \text{ mm}.

First, simplify the coefficient of $x$ :

20 \times 10^{-3} = 0.02 \, \text{mm}^{-1}.

So the equation becomes:

y(x, t) = 4.0 \sin\left(0.02x + 600t\right) \text{ mm}.

A standard form for a travelling wave is:

y(x, t) = A \sin(kx - \omega t)

which represents a wave moving in the positive $x$ -direction with speed $v = \dfrac{\omega}{k}$ .

Notice that our wave equation has the form:

\sin(0.02x + 600t)

The positive sign in front of $600t$ means we can rewrite the phase as:

0.02x + 600t = 0.02x - (-600t),

which indicates that the angular frequency $\omega$ in the standard form is effectively $-600$ .

The velocity $v$ of a wave is determined from the phase (for a constant phase, $\phi =$ constant):

k x + \omega t = \text{constant}.

Differentiating with respect to $t$ :

k \frac{dx}{dt} + \omega = 0,

which gives:

\frac{dx}{dt} = -\frac{\omega}{k}.

Substituting the values: $k = 0.02 \, \text{mm}^{-1}$ $\omega = 600 \, \text{s}^{-1}$ We have:

v = -\frac{600}{0.02} = -30000 \text{ mm/s}.

Convert the velocity from mm/s to m/s:

-30000 \, \text{mm/s} = -30 \, \text{m/s}.

Thus, the velocity of the wave is $-30 \, \text{m/s}$ . The correct answer is Option D.

Q94

Two cars A and B are moving away from each other in opposite directions. Both the cars are moving with a speed of 20 ms–1 with respect to the ground. If an observer in car A detects a frequency 2000 Hz of the sound coming from car B, what is the natural frequency of the sound source in car B ? (speed of sound in air = 340 ms–1) :-

A 2300 Hz

B 2060 Hz

C 2250 Hz

D 2150 Hz

Correct Answer

Option C

Solution

f = {{\left( {v \pm {u_0}} \right)} \over {\left( {v \pm {u_s}} \right)}}.{f_0} = {{\left( {v - 20} \right)} \over {\left( {v + 20} \right)}}.{f_0}

\Rightarrow 2000 = {{320} \over {360}}.{f_0}

\Rightarrow {{2000 \times 9} \over 8} = {f_0} = 2250\,Hz

Q95

Two strings with circular cross section and made of same material, are stretched to have same amount of tension. A transverse wave is then made to pass through both the strings. The velocity of the wave in the first string having the radius of cross section R is

v_1

, and that in the other string having radius of cross section R/2 is

v_2

. Then

\dfrac{v_2}{v_1}

=

A 8

B 4

C 2

D

\sqrt{2}

Correct Answer

Option C

Solution

To find the ratio of the velocities of transverse waves in two strings with different radii but identical materials and tension, consider the following: The wave velocity $v$ in a string is given by the formula: $v = \sqrt{\dfrac{T}{\mu}}$ where $T$ is the tension and $\mu$ is the linear mass density of the string, defined as: $\mu = \rho \pi R^2$ Here, $\rho$ is the density of the material, and $R$ is the radius of the string.

Given that both strings have the same tension $T$ and material, we compare their wave velocities $v_1$ and $v_2$ for radii $R_1 = R$ and $R_2 = \dfrac{R}{2}$ .

The velocity ratio is expressed as: $\dfrac{v_2}{v_1} = \dfrac{\sqrt{\dfrac{T}{\rho \pi R_2^2}}}{\sqrt{\dfrac{T}{\rho \pi R_1^2}}}$ Simplifying this expression: $\dfrac{v_2}{v_1} = \dfrac{\sqrt{R_1^2}}{\sqrt{R_2^2}} = \dfrac{R_1}{R_2} = \dfrac{R}{\dfrac{R}{2}} = 2$ Thus, the ratio $\dfrac{v_2}{v_1}$ is 2.

Q96

Tube

A

has bolt ends open while tube

B

has one end closed, otherwise they are identical. The ratio of fundamental frequency of tube

A

and

B

is

A

1:2

B

1:4

C

2:1

D

4:1

Correct Answer

Option C

Solution

KEY CONCEPT : The fundamental frequency for closed organ pipe is given by

{\upsilon _c} = {v \over {4\ell }}

and For open organ pipe is given by

{\upsilon _0} = {v \over {2\ell }}

$\therefore$

{{{\upsilon _0}} \over {{\upsilon _c}}} = {v \over {2\ell }} \times {{4\ell } \over v} = {2 \over 1}