Waves Questions — JEE Physics

Q1

A wire of length L and mass per unit length 6.0 × 10–3 kgm–1 is put under tension of 540 N. Two consecutive frequencies that it resonates at are : 420 Hz and 490 Hz. Then L in meters is :

A 5.1 m

B 2.1 m

C 1.1 m

D 8.1 m

Correct Answer

Option B

Solution

Fundamental frequency = 70 Hz. 70 =

{1 \over {2l}}\sqrt {{T \over \mu }}

$\Rightarrow$

l

= 2.14 m

Q2

length of a string tied to two rigid supports is

40

cm

. Maximum length (wavelength in

cm

) of a stationary wave produced on it is

A

20

B

80

C

40

D

120

Correct Answer

Option B

Solution

This will happen for fundamental mode of vibration as shown in the figure.

{S_1}

and

{S_2}

are rigid support Here

{\lambda \over 2} = 40\,\,\,\,\,\,

$\therefore$

\lambda = 80\,cm

Q3

Which of the following equations represents a travelling wave?

A y = Aexcos(

\omega

t

-

\theta

)

B y = Ae

-

x2(vt +

\theta

)

C y = A sin (15x

-

2t)

D y = A sinx cos

\omega

t

Correct Answer

Option C

Solution

Y = F(x, t) For travelling wave y should be linear function of x and t and they must exist as (x $\pm$ vt) Y = A sin (15x $-$ 2t) $\to$ linear function in x and t.

Q4

A string 2.0 m long and fixed at its ends is driven by a 240 Hz vibrator. The string vibrates in its third harmonic mode. The speed of the wave and its fundamental frequency is :-

A 180m/s, 80 Hz

B 180m/s, 120 Hz

C 320m/s, 120 Hz

D 320m/s, 80 Hz

Correct Answer

Option D

Solution

We have:

f = {{nv} \over {2l}}

240 = {{3 \times v} \over {2 \times 2}}

$\Rightarrow$ v = 320 m/s Fundamental frequency =

{v \over {2l}}

= 80 Hz.

Q5

The velocity of sound in a gas, in which two wavelengths 4.08 m and 4.16 m produce 40 beats in 12s, will be :

A 282.8 ms

-

1

B 175.5 ms

-

1

C 353.6 ms

-

1

D 707.2 ms

-

1

Correct Answer

Option D

Solution

{v \over {4.08}} - {v \over {4.16}} = {{40} \over {12}}

v = {{40} \over {12}} \times {{4.08 \times 4.16} \over {0.08}}

= 707.2

m/s

Q6

Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason

\mathbf{R}

Assertion A: A sound wave has higher speed in solids than gases. Reason R: Gases have higher value of Bulk modulus than solids. In the light of the above statements, choose the correct answer from the options given below

A Both

\mathbf{A}

and

\mathbf{R}

are true and

\mathbf{R}

is the correct explanation of

\mathbf{A}

B

\mathbf{A}

is false but

\mathbf{R}

is true

C

\mathbf{A}

is true but

\mathbf{R}

is false

D Both

\mathbf{A}

and

\mathbf{R}

are true but

\mathbf{R}

is NOT the correct explanation of

\mathbf{A}

Correct Answer

Option C

Solution

We know, speed of sound in a medium

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

where B = bulk modulus, $\rho$ = density of the medium Since, solids and liquids are much more difficult to compress than gases so they have much higher values of bulk modulus. i.e.,

{B_{solid}} > {B_{liquid}} > {B_{gas}}

Generally solids and liquids have higher mass densities ( $\rho$ ) than gases.

But corresponding increase in bulk modulus is much higher.

So,

{v_{solid}} > {v_{liquid}} > {v_{gas}}

Hence, option C is correct.

Q7

Two harmonic waves moving in the same direction superimpose to form a wave

x=\mathrm{a} \cos (1.5 \mathrm{t}) \cos (50.5 \mathrm{t})

where t is in seconds. Find the period with which they beat. (close to nearest integer)

A 1 s

B 4 s

C 2 s

D 6 s

Correct Answer

Option C

Solution

The superposition of the two harmonic waves results in the wave equation: $x = a \cos(1.5t) \cos(50.5t)$ This equation can be expanded using the trigonometric identity for the product of cosines: $x = \dfrac{a}{2} \cos[(1.5 + 50.5)t] + \dfrac{a}{2} \cos[(50.5 - 1.5)t]$ Simplifying, we find: $x = \dfrac{a}{2} \cos(52t) + \dfrac{a}{2} \cos(49t)$ This represents two waves with angular frequencies $52$ and $49$ , respectively.

To find the beat frequency, we calculate the differences in frequencies: $f_1 = \dfrac{52}{2\pi}, \quad f_2 = \dfrac{49}{2\pi}$ The beat frequency is then: $f_{\text{Beat}} = f_1 - f_2 = \dfrac{3}{2\pi} \text{ Hz}$ The period of the beats, $T_{\text{Beat}}$ , is the reciprocal of the beat frequency: $T_{\text{Beat}} = \dfrac{1}{f_{\text{Beat}}} = \dfrac{2\pi}{3} \text{ sec}$ Approximating this gives: $T_{\text{Beat}} \approx 2.09 \text{ sec} \approx 2 \text{ sec}$

Q8

Displacement of a wave is expressed as

x(t)=5 \cos \left(628 t+\dfrac{\pi}{2}\right) \mathrm{m}

. The wavelength of the wave when its velocity is

300 \mathrm{~m} / \mathrm{s}

is :

(\pi=3.14)

A 0.33 m

B 0.5 m

C 3 m

D 5 m

Correct Answer

Option C

Solution

\begin{aligned} & \mathrm{x}(\mathrm{t})=5 \cos \left[628 \mathrm{t}+\frac{\pi}{2}\right] \mathrm{m} \\ & \text{ velocity }\left(\mathrm{v}_\omega\right)=300 \mathrm{~m} / \mathrm{s} \\ & \mathrm{v}_{\mathrm{w}}=\frac{\omega}{\mathrm{K}} \\ & 300=\frac{628}{\mathrm{~K}} \Rightarrow \mathrm{~K}=\frac{628}{300} \\ & \frac{2 \pi}{\lambda}=\frac{628}{300} \Rightarrow \lambda=\frac{2 \times 3.14 \times 300}{628} \\ & \lambda=2 \mathrm{~m} \end{aligned}

Q9

In the resonance experiment, two air columns (closed at one end) of 100 cm and 120 cm long, give 15 beats per second when each one is sounding in the respective fundamental modes. The velocity of sound in the air column is:

A

370 \mathrm{~m} / \mathrm{s}

B

340 \mathrm{~m} / \mathrm{s}

C

335 \mathrm{~m} / \mathrm{s}

D

360 \mathrm{~m} / \mathrm{s}

Correct Answer

Option D

Solution

The fundamental frequency for a closed (organ) pipe can be expressed as: $f = \dfrac{v}{4\ell}$ For the first air column, with length $\ell_1$ , the frequency $f_1$ is: $f_1 = \dfrac{v}{4\ell_1}$ For the second air column, with length $\ell_2$ , the frequency $f_2$ is: $f_2 = \dfrac{v}{4\ell_2}$ The beat frequency, which is the difference in these two frequencies ( $f_1 - f_2$ ), is given as 15 beats per second: $\text{Beat} = f_1 - f_2 = \dfrac{v}{4} \left( \dfrac{1}{\ell_1} - \dfrac{1}{\ell_2} \right)$ Substitute the given lengths into the formula: $15 = \dfrac{v}{4} \left( \dfrac{1}{1} - \dfrac{1}{1.2} \right)$ Simplify the equation: $15 = \dfrac{v}{4} \left( \dfrac{0.2}{1.2} \right)$ Solve for $v$ : $v = \dfrac{15 \times 4 \times 1.2}{0.2} = 60 \times 6 = 360 \, \text{m/s}$ Thus, the velocity of sound in the air column is 360 m/s.

Q10

Three harmonic waves having equal frequency

\nu

and same intensity

{I_0}

, have phase angles 0,

{\pi \over 4}

and

- {\pi \over 4}

respectively. When they are superimposed the intensity of the resultant wave is close to :

A 5.8 I0

B 3 I0

C 0.2 I0

D I0

Correct Answer

Option A

Solution

I0 = CA2 AR = A + A

\sqrt 2

= A(1 +

\sqrt 2

) IR = C

A_R^2

$\therefore$ IR = CA2

{\left( {\sqrt 2 + 1} \right)^2}

= 5.8 I0