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Waves NEET Questions — Free MCQ Practice

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A second harmonic has to be generated in a string of length l stretched between two rigid supports. The point where the string has to be plucked and touched are :

A.

Plucked at $\frac{l}{4}$ and touch at $\frac{l}{2}$

B.

Plucked at $\frac{l}{4}$ and touch at $\frac{3 l}{4}$

C.

Plucked at $\frac{l}{2}$ and touched at $\frac{l}{4}$

D.

Plucked at $\frac{l}{2}$ and touched at $\frac{3 l}{4}$

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The tension of a stretched string is increased by 69%. In order to keep its frequency of vibration constant, its length must be increased by :

A.

20%

B.

30%

C.

$\sqrt{69} %$

D.

69%

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Explanation

$n \propto \frac{\sqrt{T}}{l} \Rightarrow l \propto \sqrt{T}$ (As n = constant)

⇒ $\frac{l_{2}}{l_{1}} = \sqrt{\frac{T_{2}}{T_{1}}} = l_{1} \sqrt{\frac{169}{100}} \Rightarrow l_{2} = 1 .3 l_{1} = l_{1} + 30 %$ of l₁

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The length of a sonometer wire tuned to a frequency of 250 Hz is 0.60 metre. The frequency of tuning fork with which the vibrating wire will be in tune when the length is made 0.40 metre is :

A.

250 Hz

B.

375 Hz

C.

256 Hz

D.

384 Hz

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Explanation

$n_{1} l_{1} = n_{2} l_{2} \Rightarrow 250 \times 0 .6 = n_{2} \times 0 .4 \Rightarrow n_{2} = 375$

$\Rightarrow n_{2} = 375 H z$

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Two uniform strings A and B made of steel are made to vibrate under the same tension. If the first overtone of A is equal to the second overtone of B and if the radius of A is twice that of B, the ratio of the lengths of the strings is -

A.

1: 2

B.

1 : 3

C.

1 : 4

D.

1 : 6

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Explanation

First overtone of string A = Second overtone of string B.

⇒ Second harmonic of A = Third harmonic of B

⇒ $n_{2} = n_{3} \Rightarrow {[2 (n_{1})]}_{A} = {[3 (n_{1})]}_{B}$ ( $∴ n_{1} = \frac{1}{2 l} \sqrt{\frac{T}{π r^{2} ρ}}$ )

⇒ $2 [\frac{1}{2 l_{A} r_{A}} \sqrt{\frac{T}{π ρ}}] = 3 [\frac{1}{2 l_{B} r_{B}} \sqrt{\frac{T}{π ρ}}]$

$\frac{l_{A}}{l_{B}} = \frac{2}{3} \frac{r_{B}}{r_{A}} \Rightarrow \frac{l_{A}}{l_{B}} = \frac{2}{3} \times \frac{r_{B}}{(2 r_{B})} = \frac{1}{3}$

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Two wires are fixed in a sonometer. Their tensions are in the ratio 8 : 1. The lengths are in the ratio 36 : 35. The diameters are in the ratio 4 : 1. Densities of the materials are in the ratio 1 : 2. If the lower frequency in the setting is 360 Hz. the beat frequency when the two wires are sounded together is :

A.

5

B.

8

C.

6

D.

10

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Explanation

Frequency in a stretched string is given by $n = \frac{1}{2 l} \sqrt{\frac{T}{π r^{2} ρ}} = \frac{1}{l} \sqrt{\frac{T}{π d^{2} ρ}}$ (d = Diameter of string)

⇒ $\frac{n_{1}}{n_{2}} = \frac{l_{2}}{l_{1}} \sqrt{\frac{T_{1}}{T_{2}} \times {(\frac{d_{2}}{d_{1}})}^{2} \times (\frac{ρ_{2}}{ρ_{1}})}$

$= \frac{35}{36} \sqrt{\frac{8}{1} \times {(\frac{1}{4})}^{2} \times \frac{2}{1}} = \frac{35}{36}$

$\Rightarrow n_{2} = \frac{36}{35} \times 360 = 370$

Hence beat frequency = $n_{2} - n_{1} = 10$

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The first overtone of a stretched wire of given length is 320 Hz. The first harmonic is :

A.

320 Hz

B.

160 Hz

C.

480 Hz

D.

640 Hz

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Explanation

Frequency of first overtone or second harmonic (n₂) = 320 Hz. So, frequency of first harmonic $n_{1} = \frac{n_{2}}{2} = \frac{320}{2} = 160 H z$

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The sound carried by the air from a sitar to a listener is a wave of the following type :

A.

Longitudinal stationery

B.

Transverse progressive

C.

Transverse stationery

D.

Longitudinal progressive

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Explanation

Observer receives sound waves (music) which are longitudinal progressive waves.

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Three similar wires of frequency n₁, n₂ and n₃ are joined to make one wire. Its frequency will be :

A.

$n = n_{1} + n_{2} + n_{3}$

B.

$\frac{1}{n} = \frac{1}{n_{1}} + \frac{1}{n_{2}} + \frac{1}{n_{3}}$

C.

$\frac{1}{\sqrt{n}} = \frac{1}{\sqrt{n_{1}}} + \frac{1}{\sqrt{n_{2}}} + \frac{1}{\sqrt{n_{3}}}$

D.

$\frac{1}{n^{1}} = \frac{1}{n_{1}^{2}} + \frac{1}{n_{2}^{2}} + \frac{1}{n_{3}^{2}}$

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Explanation

$n = \frac{1}{2 l} \sqrt{\frac{T}{m}}$ ⇒ $n_{1} l_{1} = n_{2} l_{2} = n_{3} l_{3} = k$

$l_{1} + l_{2} + l_{3} = l$ ⇒ $\frac{k}{n_{1}} + \frac{k}{n_{2}} + \frac{k}{n_{3}} = \frac{k}{n}$

⇒ $\frac{1}{n} = \frac{1}{n_{1}} + \frac{1}{n_{2}} + \frac{1}{n_{3}} + . .......$

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Two vibrating strings of the same material but lengths L and 2L have radii 2r and r respectively. They are stretched under the same tension. Both the strings vibrate in their fundamental modes, the one of length L with frequency n₁ and the other with frequency n₂. The ratio n₁/n₂ is given by :

A.

2

B.

4

C.

8

D.

1

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Explanation

Fundamental frequency $n = \frac{1}{2 l} \sqrt{\frac{T}{π r^{2} ρ}}$

where m = Mass per unit length of wire

⇒ $n \propto = \frac{1}{l r} \Rightarrow \frac{n_{1}}{n_{2}} = \frac{r_{2}}{r_{1}} \times \frac{l_{2}}{l_{1}} = \frac{r}{2 r} \times \frac{2 L}{L} = \frac{1}{1}$

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A string is rigidly tied at two ends and its equation of vibration is given by $y = (\cos 2 π t) (\sin 2 π x)$ Then minimum length of the string is :

A.

1 m

B.

$\frac{1}{2} m$

C.

5 m

D.

2π m

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Explanation

Given equation of stationary wave is

$y = \sin 2 π x \cos 2 π t$ ,

comparing it with standard equation $y = 2 A \sin \frac{2 π x}{λ} \cos ω t$

We have $\frac{2 π x}{λ} = 2 π x$ ⇒ $λ = 1 m$

Minimum length of string (first mode) , $L_{\min} = \frac{λ}{2} = \frac{1}{2} m$

Waves MCQs for NEET — Physics Questions with Answers

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