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Waves MCQs for NEET — Physics Questions with Answers

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A string of length 2 m is fixed at both ends. If this string vibrates in its fourth normal mode with a frequency of 500 Hz then the waves would travel on its with a velocity of :

125 m/s

250 m/s

500 m/s

1000 m/s

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Explanation

For string $λ = \frac{2 l}{p}$

where p = No. of loops = Order of vibration

Hence for forth mode p = 4 ⇒ $λ = \frac{l}{2}$

Hence v = nλ $= 500 \times \frac{2}{2} = 500 H z$

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The fundamental frequency of a sonometre wire is n. If its radius is doubled and its tension becomes half, the material of the wire remains same, the new fundamental frequency will be :

$\frac{n}{\sqrt{2}}$

$\frac{n}{2}$

$\frac{n}{2 \sqrt{2}}$

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Explanation

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

⇒ $\frac{n_{2}}{n_{1}} = \frac{r_{1}}{r_{2}} \sqrt{\frac{T_{2}}{T_{1}}}$ $= \frac{1}{2} \times \sqrt{\frac{1}{2}} = \frac{1}{2 \sqrt{2}}$

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In an experiment with a sonometer, a tuning fork of frequency 256 Hz resonates with a length of 25 cm and another tuning fork resonates with a length of 16 cm. The tension of the string remaining constant the frequency of the second tuning fork is :

163.84 Hz

400 Hz

320 Hz

204.8 Hz

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Explanation

In case of sonometer frequency is given by

$n = \frac{p}{2 l} \sqrt{\frac{T}{m}}$ ⇒ $\frac{n_{2}}{n_{1}} = \frac{l_{1}}{l_{2}} \Rightarrow n_{2} = \frac{25}{16} \times 256 = 400 H z$

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The length of two open organ pipes are l and $(l + Δ l)$ respectively. Neglecting end correction, the frequency of beats between them will be approximate :

$\frac{v}{2 l}$

$\frac{v}{4 l}$

$\frac{v Δ l}{2 l^{2}}$

$\frac{v Δ l}{l}$

(Here v is the speed of sound)

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Explanation

$λ_{1} = 2 l, λ_{2} = 2 l + 2 Δ l$ ⇒ $n_{1} = \frac{v}{2 l}$ and $n_{2} = \frac{v}{2 l + 2 Δ l}$

⇒ No. of beats $= n_{1} - n_{2} = \frac{v}{2} (\frac{1}{l} - \frac{1}{l + Δ l}) = \frac{v Δ l}{2 l^{2}}$

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A tube closed at one end and containing air is excited. It produces the fundamental note of frequency of 512 Hz. If the same tube is open at both the ends the fundamental frequency that can be produced is :

1024 Hz

512 Hz

256 Hz

128 Hz

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Explanation

The fundamental frequency of open pipe is double that of the closed pipe.

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A closed pipe and an open pipe have their first overtones identical in frequency. Their lengths are in the ratio :

1 : 2

2 : 3

3 : 4

4 : 5

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Explanation

It is given that the first overtone of closed pipe = First overtone of the open pipe ⇒ $3 (\frac{v}{4 l_{1}}) = 2 (\frac{v}{2 l_{2}})$ ; where l₁ and l₂ are the lengths of closed and open organ pipes hence $\frac{l_{1}}{l_{2}} = \frac{3}{4}$ .

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An empty vessel is getting filled with water, then the frequency of vibration of the air column in the vessel

Remains the same

Decreases

Increases

First increases then decrease

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Explanation

For closed pipe in general $n = \frac{v}{4 l} (2 N - 1)$ ⇒ $n \propto \frac{1}{l}$

i.e. if length of air column decreases, frequency increases.

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If the velocity of sound in air is 350 m/s. Then the fundamental frequency of an open organ pipe of length 50 cm, will be

350 Hz

175 Hz

900 Hz

750 Hz

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Explanation

Fundamental frequency of open pipe $n_{1} = \frac{v}{2 l} = \frac{350}{2 \times 0 .5} = 350 H z$ .

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The fundamental note produced by a closed organ pipe is of frequency f. The fundamental note produced by an open organ pipe of same length will be of frequency

$\frac{f}{2}$

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Explanation

$n_{closed} = \frac{v}{4 l}, n_{open} = \frac{v}{2 l}$ ⇒ $n_{open} = 2 n_{closed} = 2 f$

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If the velocity of sound in air is 336 m/s. The maximum length of a closed pipe that would produce a just audible sound will be :

3.2 cm

4.2 m

4.2 cm

3.2 m

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Explanation

Minimum audible frequency = 20 Hz.

$\Rightarrow \frac{v}{4 l} = 20 \Rightarrow l = \frac{336}{4 \times 20} = 4 .2 m$

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