Waves — Page 7 | JEE Physics

Q61

The fundamental frequency of a closed organ pipe is equal to the first overtone frequency of an open organ pipe. If length of the open pipe is

60 \mathrm{~cm}

, the length of the closed pipe will be:

A 15 cm

B 60 cm

C 45 cm

D 30 cm

Correct Answer

Option A

Solution

\begin{aligned} & \mathrm{f}_1=\frac{\mathrm{v}}{4 \mathrm{~L}_1} \\ & \mathrm{f}_1=\mathrm{f}_2 \\ & \frac{\mathrm{v}}{4 \mathrm{~L}_1}=\frac{\mathrm{v}}{\mathrm{L}_2} \\ & \Rightarrow \mathrm{L}_2=4 \mathrm{~L}_1 \\ & 60=4 \times \mathrm{L}_1 \\ & \mathrm{~L}_1=15 \mathrm{~cm} \end{aligned}

Q62

A plane progressive wave is given by

y=2 \cos 2 \pi(330 \mathrm{t}-x) \mathrm{m}

. The frequency of the wave is :

A 660 Hz

B 340 Hz

C 330 Hz

D 165 Hz

Correct Answer

Option C

Solution

To find the frequency of the plane progressive wave given by the equation

y = 2 \cos 2 \pi(330 \mathrm{t} - x) \mathrm{m}

, we start by analyzing the general form of a wave equation. The general form of a wave equation is:

y = A \cos (2 \pi ft - kx + \phi)

where:

A

is the amplitude of the wave.

f

is the frequency of the wave.

t

is the time variable.

k

is the wave number, defined as

\frac{2 \pi}{\lambda}

where $\lambda$ is the wavelength.

x

is the spatial variable. $\phi$ is the phase constant.

By comparing the given wave equation with the general form, we have:

y = 2 \cos 2 \pi (330 t - x)

We observe that the term

2 \pi(330t - x)

corresponds to

2 \pi ft - kx

in the general form. From this, it is clear that:

2 \pi ft \rightarrow 2 \pi \cdot 330 t

Therefore,

f = 330

Hz So, the frequency of the wave is 330 Hz. Thus, the correct answer is: Option C: 330 Hz

Q63

A closed organ and an open organ tube are filled by two different gases having same bulk modulus but different densities

\rho_1

and

\rho_2

, respectively. The frequency of

9^{\text{th }}

harmonic of closed tube is identical with

4^{\text{th }}

harmonic of open tube. If the length of the closed tube is 10 cm and the density ratio of the gases is

\rho_1: \rho_2=1: 16

, then the length of the open tube is :

A

\dfrac{15}{7} \mathrm{~cm}

B

\dfrac{20}{9} \mathrm{~cm}

C

\dfrac{20}{7} \mathrm{~cm}

D

\dfrac{15}{9} \mathrm{~cm}

Correct Answer

Option B

Solution

We know, for closed pipe,

{f_n} = {{nv} \over {4l}},\,n = 1,3,5,7

for open pipe,

{f_n} = {{nv} \over {2L}},\,n = 1,2,3,4

So, 9 $^{th}$ harmonic of closed pipe

= {{9{v_1}} \over {4{l_1}}}

4 $^{th}$ harmonic of open pipe

= {{4{v_2}} \over {2{l_2}}} = {{2{v_2}} \over {{l_2}}}

$\therefore$

{{9{v_1}} \over {4{l_1}}} = {{2{v_2}} \over {{l_2}}} \Rightarrow {{{l_2}} \over {{l_1}}} = {8 \over 9}{{{v_2}} \over {{v_1}}}

We know,

v = \sqrt {{\beta \over \rho }}

So,

{{{v_2}} \over {{v_1}}} = \sqrt {{{{\rho _1}} \over {{\rho _2}}}}

(for same $\beta$ ) hence,

{{{l_2}} \over {{l_1}}} = {8 \over 9}\sqrt {{{{\rho _1}} \over {{\rho _2}}}} = {8 \over 9}\sqrt {{1 \over {16}}} = {8 \over 9} \times {1 \over 4}

\Rightarrow {l_2} = {2 \over 9} \times {l_1} \Rightarrow {l_2} = {2 \over 9} \times 10 = {{20} \over 9}cm

Q64

The amplitude and phase of a wave that is formed by the superposition of two harmonic travelling waves,

y_1(x, t) = 4 \sin (kx - \omega t)

and

y_2(x, t) = 2 \sin (kx - \omega t + \dfrac{2\pi}{3})

, are: (Take the angular frequency of initial waves same as

\omega

)

A

\left[\sqrt{3}, \dfrac{\pi}{6}\right]

B

\left[2\sqrt{3}, \dfrac{\pi}{6}\right]

C

\left[6, \dfrac{2\pi}{3}\right]

D

\left[6, \dfrac{\pi}{3}\right]

Correct Answer

Option B

Solution

To determine the amplitude and phase of the wave formed by the superposition of the two harmonic traveling waves, $y_1(x, t) = 4 \sin (kx - \omega t)$ and $y_2(x, t) = 2 \sin (kx - \omega t + \dfrac{2\pi}{3})$ , we use the following steps: First, calculate the resultant amplitude $A$ using the formula for the resultant of two waves with different phases: $A = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 \cos(\phi)}$ where: $A_1 = 4$ (amplitude of the first wave) $A_2 = 2$ (amplitude of the second wave) $\phi = 120^\circ$ (phase difference, given by $\dfrac{2\pi}{3}$ ).

Substituting the values, we get: $A = \sqrt{4^2 + 2^2 + 2 \times 4 \times 2 \times \cos(120^\circ)}$ $\cos(120^\circ) = -\dfrac{1}{2}$ Therefore: $A = \sqrt{16 + 4 + 16 \times (-\dfrac{1}{2})}$ $A = \sqrt{16 + 4 - 8}$ $A = \sqrt{12} = 2\sqrt{3}$ Next, compute the phase $\phi$ of the resultant wave: $\tan \phi = \dfrac{A_2 \sin(\phi)}{A_1 + A_2 \cos(\phi)}$ $\tan \phi = \dfrac{2 \sin(120^\circ)}{4 + 2 \cos(120^\circ)}$ $\sin(120^\circ) = \dfrac{\sqrt{3}}{2} \quad \text{and} \quad \cos(120^\circ) = -\dfrac{1}{2}$ Substituting these values, we get: $\tan \phi = \dfrac{2 \times \dfrac{\sqrt{3}}{2}}{4 + 2 \times -\dfrac{1}{2}}$ $\tan \phi = \dfrac{\sqrt{3}}{3}$ This simplifies to: $\phi = \dfrac{\pi}{6}$ Thus, the amplitude and phase of the resultant wave are $2\sqrt{3}$ and $\dfrac{\pi}{6}$ respectively.

Q65

A sinusoidal wave of wavelength 7.5 cm travels a distance of 1.2 cm along the

x

-direction in 0.3 sec . The crest P is at

x=0

at

\mathrm{t}=0 \mathrm{sec}

and maximum displacement of the wave is 2 cm . Which equation correctly represents this wave?

A

y=2 \cos (0.83 x-3.35 t) \mathrm{cm}

B

y=2 \sin (0.83 x-3.5 \mathrm{t}) \mathrm{cm}

C

y=2 \cos (0.13 x-0.5 t) \mathrm{cm}

D

y=2 \cos (3.35 x-0.83 \mathrm{t}) \mathrm{cm}

Correct Answer

Option A

Solution

To find the equation of the sinusoidal wave, we need to determine several properties of the wave: Wave Velocity (v): Given that the wave travels a distance of 1.2 cm in 0.3 seconds, the velocity $v$ can be computed as: $v = \dfrac{\text{distance}}{\text{time}} = \dfrac{1.2 \, \text{cm}}{0.3 \, \text{s}} = 4 \, \text{cm/s}$ Wave Number (k): The wave number $k$ is calculated using the wavelength $\lambda = 7.5 \, \text{cm}$ : $k = \dfrac{2\pi}{\lambda} = \dfrac{2\pi}{7.5} = \dfrac{4\pi}{15} \approx 0.83$ Angular Frequency ( $\omega$ ): Angular frequency $\omega$ is related to the wave number and velocity by the equation $v = \dfrac{\omega}{k}$ , thus: $\omega = vk = 4 \times \dfrac{4\pi}{15} = \dfrac{16\pi}{15} \approx 3.35$ Wave Equation: The general form for a sinusoidal wave traveling in the positive x-direction is: $y = A \cos(kx - \omega t)$ Given the maximum displacement (amplitude $A$ ) of the wave is 2 cm, the equation becomes: $y = 2 \cos(0.83x - 3.35t) \, \text{cm}$ This equation accurately describes the sinusoidal wave given the provided parameters.

Q66

In an experiment with a closed organ pipe, it is filled with water by

\left(\dfrac{1}{5}\right)

th of its volume. The frequency of the fundamental note will change by

A

20 \%

B

25 \%

C

-20 \%

D

-25 \%

Correct Answer

Option B

Solution

\begin{aligned} &\lambda_1=4 \ell\\ &\mathrm{f}_1=\frac{\mathrm{v}}{4 \ell} \end{aligned}

\begin{aligned} & \lambda_2=\frac{16 \ell}{5} \\ & \mathrm{f}_2=\frac{5 \mathrm{~V}}{16 \ell} \\ & \frac{\Delta \mathrm{f}}{\mathrm{f}}=\frac{\frac{\mathrm{V}}{\ell}\left(\frac{1}{16}\right)}{\frac{\mathrm{V}}{4 \ell}} \times 100=25 \% \end{aligned}

Q67

A wave travelling along the

x

-axis is described by the equation

y(x, t)=0.005

\cos \,\left( {\alpha \,x - \beta t} \right).

If the wavelength and the time period of the wave are

0.08

m

and

2.0s

, respectively, then

\alpha

and

\beta

in appropriate units are

A

\alpha = 25.00\pi ,\,\beta = \pi

B

\alpha = {{0.08} \over \pi },\,\beta = {{2.0} \over \pi }

C

\alpha = {{0.04} \over \pi },\,\beta = {{1.0} \over \pi }

D

\alpha = 12.50\pi ,\,\beta = {\pi \over {2.0}}

Correct Answer

Option A

Solution

y\left( {x,t} \right) = 0.005\,\cos \left( {\alpha x - \beta t} \right)

(Given) Comparing it with the standard equation of wave

y\left( {x,t} \right) = a\cos \left( {kx - \omega t} \right)

we get

k = \alpha

\,\,\,\,\,

and

\,\,\,\,\,

\omega = \beta

$\therefore$

{{2\pi } \over \gamma } = \alpha

\,\,\,\,\,

and

\,\,\,\,\,

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

$\therefore$

\alpha = {{2\pi } \over {0.08}} = 25\pi

\,\,\,\,\,

and

\,\,\,\,\,

\beta = {{2\pi } \over 2} = \pi

Q68

The equation of a wave travelling on a string is y = sin[20πx + 10πt], where x and t are distance and time in SI units. The minimum distance between two points having the same oscillating speed is :

A 10 cm

B 2.5 cm

C 20 cm

D 5.0 cm

Correct Answer

Option D

Solution

The equation for a wave traveling on a string is given by $y = \sin(20\pi x + 10\pi t)$ , where $x$ is the distance and $t$ is the time in SI units.

To find the minimum distance between two points having the same oscillating speed, we use the concept that this distance is half the wavelength $\left(\dfrac{\lambda}{2}\right)$ .

To find the wavelength $\lambda$ : $\lambda = \dfrac{2\pi}{k}$ Here, the wave number $k = 20\pi$ .

Thus, $\lambda = \dfrac{2\pi}{20\pi} = \dfrac{1}{10} \text{ m} = 10 \text{ cm}$ Thus, the minimum distance between two points having the same speed is: $\text{Distance} = \dfrac{\lambda}{2} = \dfrac{10 \text{ cm}}{2} = 5 \text{ cm}$

Q69

A heavy ball of mass M is suspendeed from the ceiling of a car by a light string of mass m (m < < M). When the car is at rest, the speed of transverse waves in the string is 60 ms

-

1. When the car has acceleration a, the wave-speed increases to 60.5 ms

-

1. The value of a, in terms of gravitational acceleration g, is closest to :

A

{g \over {30}}

B

{g \over 5}

C

{g \over 10}

D

{g \over 20}

Correct Answer

Option B

Solution

Resultant force on the ball of mass M when car is moving with a acceleration a is , Fnet =

\sqrt {{{\left( {Mg} \right)}^2} + {{\left( {Ma} \right)}^2}}

=

M\sqrt {{g^2} + {a^2}}

$\therefore$ T = M

\sqrt {{g^2} + {a^2}}

We know, Velocity, V =

\sqrt {{T \over \mu }}

When Car is at rest then, 60 =

\sqrt {{{Mg} \over \mu }}

. . . . (1) and when is moving then 60.5 =

\sqrt {{{M\sqrt {{g^2} + {a^2}} } \over \mu }}

. . . . (2) By dividing (2) by (1) we get,

{{60.5} \over {60}} = \sqrt {{{\sqrt {{g^2} + {a^2}} } \over g}}

$\Rightarrow$

\left( {1 + {{0.5} \over {60}}} \right)

=

{\left( {{{{g^2} + {a^2}} \over {{g^2}}}} \right)^{{1 \over 4}}}

$\Rightarrow$

{{{g^2} + {a^2}} \over {{g^2}}}

=

{\left( {1 + {{0.5} \over {60}}} \right)^4}

$\Rightarrow$

{{{g^2} + {a^2}} \over {{g^2}}}

= 1 + 4 $\times$

{{{0.5} \over {60}}}

[Using Binomial approximation] $\Rightarrow$

{{{g^2} + {a^2}} \over {{g^2}}}

= 1 +

{1 \over {30}}

$\Rightarrow$ 1 +

{{{a^2}} \over {{g^2}}}

= 1 +

{1 \over {30}}

$\Rightarrow$

{a \over g}

=

{1 \over {\sqrt {30} }}

$\Rightarrow$ a =

{g \over {\sqrt {30} }}

$\therefore$ Closest answer, a =

{g \over 5}

Q70

A tuning fork A of unknown frequency produces 5 beats/s with a fork of known frequency 340 Hz. When fork A is filed, the beat frequency decreases to 2 beats/s. What is the frequency of fork A?

A 335 Hz

B 345 Hz

C 338 Hz

D 342 Hz

Correct Answer

Option A

Solution

Initially beat frequency = 5Hz so, $\rho$ A = 340 $\pm$ 5 = 345 Hz, or 335 Hz after filing frequency increases slightly so, new value of frequency of A > $\rho$ A Now, beat frequency = 2Hz $\Rightarrow$ new $\rho$ A = 340 $\pm$ 2 = 342 Hz, or 338 Hz hence, original frequency of A is $\rho$ A = 335 Hz