Waves Questions — NEET Physics

Q1

For a travelling harmonic wave

y(x, t)=2.0 \cos 2 \pi(10 t-0.0080 x+0.35)

, where

x

and

y

are in cm and

t

in

s

. The phase difference between oscillatory motion of two points separated by a distance of 0.5 m is:

A

0.08 \pi \mathrm{rad}

B

\quad 0.8 \pi \mathrm{rad}

C

8 \pi \mathrm{rad}

D

0.008 \pi \mathrm{rad}

Correct Answer

Option B

Solution

$y(x, t)=2.0 \cos 2 \pi(10 t-0.008 x+0.35)$ Total phase

\begin{aligned} & \phi=20 \pi-2 \pi \times 8 \times 10^{-3} x+2 \pi \times 0.35 \\ & \Delta \phi=k \Delta x \\ & \Delta \phi=2 \pi \times 8 \times 10^{-3} \Delta x \\ & =2 \pi \times 8 \times 10^{-3} \times\left(\frac{100}{2}\right) \\ & =8 \pi \times 10^{-1}=0.8 \pi \end{aligned}

Q2

A pipe open at both ends has a fundamental frequency

f

in air. The pipe is now dipped vertically in a water drum to half of its length. The fundamental frequency of the air column is now equal to:

A

\dfrac{3 f}{2}

B

2 f

C

\dfrac{f}{2}

D

f

Correct Answer

Option D

Solution

Fundamental frequency of open pipe (at both ends) $f=\dfrac{V}{2 L} \ldots$ (i) Now immersed in water open pipe behaves as closed pipe.

\begin{aligned} & f^{\prime}=\frac{v}{4\left(\frac{L}{2}\right)}=\frac{v}{2 L} \quad\text{..... (ii)}\\ & f=f^{\prime} \end{aligned}

Q3

The displacement of a travelling wave

y=C \sin \dfrac{2 \pi}{\lambda}

(at

-x

) where

t

is time,

x

is distance and

\lambda

is the wavelength, all in S.I. units. Then the frequency of the wave is

A

\dfrac{2 \pi \lambda}{a}

B

\dfrac{2 \pi a}{\lambda}

C

\dfrac{\lambda}{a}

D

\dfrac{a}{\lambda}

Correct Answer

Option D

Solution

To find the frequency of the wave, we need to start by analyzing the given displacement equation of the wave:

y = C \sin \left( \frac{2 \pi}{\lambda} (at - x) \right)

Where:

y

is the displacement of the wave

C

is the amplitude

\frac{2 \pi}{\lambda}

is the wave number (denoting how many wavelengths fit into a unit length) $\lambda$ is the wavelength

a

is some constant (likely representing the speed of the wave)

t

is time

x

is the distance By comparing with the standard form of a travelling wave, we have:

y = C \sin \left( k (at - x) \right)

Where

k

is the wave number:

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

From the standard wave equation, the argument of the sine function is usually written as:

k (at - x)

This implies that

ka

in the wave equation represents the angular frequency $\omega$ of the wave:

\omega = k a

Substituting

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

into $\omega$ :

\omega = \left( \frac{2 \pi}{\lambda} \right) a = \frac{2 \pi a}{\lambda}

Angular frequency $\omega$ is related to the frequency

f

by:

\omega = 2 \pi f

So:

2 \pi f = \frac{2 \pi a}{\lambda}

Solving for the frequency

f

:

f = \frac{a}{\lambda}

Therefore, the correct answer is: Option D:

\frac{a}{\lambda}

Q4

The

4^{\text{th }}

overtone of a closed organ pipe is same as that of

3^{\text{rd }}

overtone of an open pipe. The ratio of the length of the closed pipe to the length of the open pipe is :

A 8 : 9

B 9 : 7

C 9 : 8

D 7 : 9

Correct Answer

Option C

Solution

\begin{aligned} & \mathrm{n}_{\text{oop }}=(2 \mathrm{M}+1)^{\text{th }} \text{ Har. }=(2 \times 4+1) \times \frac{\mathrm{V}}{4 \ell_{\mathrm{c}}}=\frac{9 \mathrm{~V}}{4 \ell_{\mathrm{c}}} \\ & \mathrm{n}_{\text{oop }}=(\mathrm{M}+1)^{\text{th }} \text{ Har. }=(3+1) \frac{\mathrm{V}}{2 \ell_0}=\frac{4 \mathrm{~V}}{2 \ell_0} \\ & \text{ now } \frac{9 \mathrm{~V}}{4 \ell_{\mathrm{c}}}=\frac{4 \mathrm{~V}}{2 \ell_0} \\ & \frac{\ell_{\mathrm{c}}}{\ell_0}=\frac{18}{16}=\frac{9}{8} \end{aligned}

Q5

The ratio of frequencies of fundamental harmonic produced by an open pipe to that of closed pipe having the same length is

A

2: 1

B

1: 3

C

3: 1

D

1: 2

Correct Answer

Option A

Solution

The fundamental frequency of an open pipe, n oop , is given by V / (2L) , where V is the speed of sound and L is the length of the pipe.

For a closed pipe of the same length, the fundamental frequency, n cop , is V / (4L) since only half as many wavelengths fit into the same length due to the closed end.

Therefore, the ratio of frequencies of an open pipe to a closed pipe is:

\frac{n_{\text{oop}}}{n_{\text{cop}}} = \frac{V / (2L)}{V / (4L)} = \frac{4L}{2L} = 2:1

This shows that the fundamental frequency of an open pipe is twice that of a closed pipe of the same length.

Q6

An organ pipe filled with a gas at 27

^\circ

C resonates at 400 Hz in its fundamental mode. If it is filled with the same gas at 90

^\circ

C, the resonance frequency at the same mode will be

A 512 Hz

B 420 Hz

C 440 Hz

D 484 Hz

Correct Answer

Option C

Solution

{f_0} = {v \over {2L}} = {{\sqrt {{{\gamma RT} \over M}} } \over {2L}}

{{{f_{{0_1}}}} \over {{f_{{0_2}}}}} = \sqrt {{{{T_1}} \over {{T_2}}}} = \sqrt {{{300} \over {363}}} = {1 \over {1.1}}

{f_{{0_2}}} = 1\,.\,1{f_{{0_1}}} = 1\,.\,1(400)

= 440

Hz

Q7

If the initial tension on a stretched string is doubled, then the ratio of the initial and final speeds of a transverse wave along the string is

A 1 : 1

B

\sqrt2

: 1

C 1 :

\sqrt2

D 1 : 2

Correct Answer

Option C

Solution

We know, velocity of transverse wave

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

$\therefore$

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

and

{v_f} = \sqrt {{{2T} \over \mu }}

$\therefore$

{{{v_i}} \over {{v_f}}} = {1 \over {\sqrt 2 }}

Q8

In a guitar, two strings A and B made of same material are slightly out of tune and produce beats of frequency 6 Hz. When tension in B is slightly decreased, the beat frequency of A is 530 Hz, the original freqnency of B will be :

A 524 Hz

B 536 Hz

C 537 Hz

D 523 Hz

Correct Answer

Option A

Solution

Difference of

{f _A} > {f _B}

{f_A} - {f_B}

= 6 Hz

{f_A}

= 530 Hz

{f_B}

= 524 Hz (original)

Q9

A tuning fork is used to produce resonance in a glass tube. The length of the air column in this tube can be adjusted by a variable piston. At room temperature of 27°C two successive resonances are produced at 20 cm and 73 cm of column length. If the frequency of the tuning fork is 320 Hz, the velocity of sound in air at 27°C is

A 330 m s –1

B 339 m s –1

C 350 m s –1

D 300 m s –1

Correct Answer

Option B

Solution

Two successive resonance are produced at 20 cm and 73 cm of column length $\therefore$

{\lambda \over 2}

= (73 - 20) $\times$ 10 -2 m $\Rightarrow$ $\lambda$ = 2 $\times$ (73 - 20) $\times$ 10 -2 m Velocity of sound, v = n $\lambda$ = 2 $\times$ 320 $\times$ (73 - 20) $\times$ 10 -2 m = 339 m/s

Q10

The fundamental frequency in an open organ pipe is equal to the third harmonic of a closed organ pipe. If the length of the closed organ pipe is 20 cm, the length of the open organ pipe is

A 13.2 cm

B 8 cm

C 12.5 cm

D 16 cm

Correct Answer

Option A

Solution

For closed organ pipe, third harmonic n =

{{\left( {2N - 1} \right)V} \over {4l}} = {{3V} \over {4l}}

( $\because$ N = 2 ) For open organ pipe, fundamental frequenty n =

{{NV} \over {2l}} = {V \over {2l'}}

( $\because$ N = 1 ) Given, third harmonic for closed organ pipe = fundamental frequency for open organ pipe.

$\therefore$

{{3V} \over {4l}} = {V \over {2l'}}

$\Rightarrow$ l' =

{{4l} \over {3 \times 2}}

=

{{2l} \over 3}

=

{{2 \times 20} \over 3}

= 13.33 cm