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41 forks are so arranged that each produces 5 beats per sec when sounded with its near fork. If the frequency of the last fork is double the frequency of the first fork, then the frequencies of the first and last fork are respectively :

200, 400

205, 410

195, 390

100, 200

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Explanation

Similar to previous question

n_First = n_First + (N – 1)x

2n = n + (41 – 1) × 5

⇒ n_First = 200 Hz and n_Last = 400 Hz

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Two identical wires have the same fundamental frequency of 400 Hz when kept under the same tension. If the tension in one wire is increased by 2%, the number of beats produced will be :

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Explanation

$n \propto \sqrt{T}$ ⇒ $\frac{Δ n}{n} = \frac{1}{2} \frac{Δ T}{T}$

Beat frequency $= Δ n = (\frac{1}{2} \frac{Δ T}{T}) n = \frac{1}{2} \times \frac{2}{100} \times 400 = 4$

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16 tunning forks are arranged in the order of increasing frequencies. Any two successive forks give 8 beats per sec when sounded together. If the frequency of the last fork is twice the first, then the frequency of the first fork is

120

160

180

220

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Explanation

Using n_Last = n_First + (N – 1)x

⇒ 2n = n + (16 – 1) × 8 ⇒ n = 120 Hz

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The frequency of a stretched uniform wire under tension is in resonance with the fundamental frequency of a closed tube. If the tension in the wire is increased by 8 N, it is in resonance with the first overtone of the closed tube. The initial tension in the wire is

1 N

4 N

8 N

16 N

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Explanation

According to problem

$\frac{1}{2 L} \sqrt{\frac{T}{m}} = \frac{v}{4 L}$ …..(i)

and $\frac{1}{2 L} \sqrt{\frac{T + 8}{m}} = \frac{3 v}{4 L}$ ..…(ii)

Dividing equation (i) and (ii), $\sqrt{\frac{T}{T + 8}} = \frac{1}{3} \Rightarrow T = 1 N$

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A metal wire of linear mass density of 9.8 g/m is stretched with a tension of 10 kg weight between two rigid supports 1 metre apart. The wire passes at its middle point between the poles of a permanent magnet, and it vibrates in resonance when carrying an alternating current of frequency n. The frequency n of the alternating source is :

25 Hz

50 Hz

100 Hz

200 Hz

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Explanation

In condition of resonance, frequency of a.c. will be equal to natural frequency of wire

$n = \frac{1}{2 l} \sqrt{\frac{T}{m}} = \frac{1}{2 \times 1} \sqrt{\frac{10 \times 9 .8}{9 .8 \times 10^{- 3}}} = \frac{100}{2} = 50 H z$

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An open pipe is in resonance in its 2^nd harmonic with tuning fork of frequency f₁. Now it is closed at one end. If the frequency of the tuning fork is increased slowly from f₁ , then again a resonance is obtained with a frequency f₂. If in this case the pipe vibrates in $n^{t h}$ harmonic, then -

n = 3, $f_{2} = \frac{3}{4} f_{1}$

n = 3, $f_{2} = \frac{5}{4} f_{1}$

n = 5, $f_{2} = \frac{5}{4} f_{1}$

n = 5, $f_{2} = \frac{3}{4} f_{1}$

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Explanation

Open pipe resonance frequency $f_{1} = \frac{2 v}{2 L}$

Closed pipe resonance frequency $f_{2} = \frac{n v}{4 L}$

$f_{2} = \frac{n}{4} f_{1}$ (where n is odd and $f_{2} > f_{1}$ )

∴ n = 5

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A string of length L and mass M hangs freely from a fixed point. Then the velocity of transverse waves along the string at a distance x from the free end is

$\sqrt{g L}$

$\sqrt{g x}$

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Explanation

The velocity of transverse waves along a stretched string depends on the tension in the string. The tension at a distance x from the free end is given by T = Mgx/L. Therefore, the velocity v = sqrt(T/Î¼) = sqrt(gx), where Î¼ is the mass per unit length of the string.

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Three waves of equal frequency having amplitudes 10 μm, 4 μm and 7 μm arrive at a given point with a successive phase difference of $\frac{π}{2}$ . The amplitude of the resulting wave in μm is given by

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Explanation

Wave 1 and 3 reach out of phase. Hence resultant phase difference between them is π.

∴ Resultant amplitude of 1 and 3 = 10 – 7 = 3 $μ m$

This wave has a phase difference of $\frac{π}{2}$ with 4 $μ m$

∴ Resultant amplitude = $\sqrt{3^{2} + 4^{2}} = 5 μ m$

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An organ pipe is closed at one end has a fundamental frequency of 1500 Hz. The maximum number of overtones generated by this pipe which a normal person can hear is :

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Explanation

The critical hearing frequency for a person is 20,000Hz.

If a closed pipe vibrate in N^th mode, then the frequency of vibration $n = \frac{(2 N - 1) v}{4 l} = (2 N - 1) n_{1}$

(where n₁ = fundamental frequency of vibration)

Hence, 20,000 $= (2 N - 1) \times 1500$ ⇒ $N = 7 .1 \approx 7$

Also, in a closed pipe;

Number of overtones = (No. of the mode of vibration) – 1

= 7 – 1 = 6.

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A transverse wave is incident on a rigid boundary. What is the phase change experienced by the reflected wave?

A. $0$ B. $\frac{\pi}{2}$ C. $\pi$ D. $2\pi$

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Explanation

According to the provided text, 'A travelling wave, at a rigid boundary or a closed end, is reflected with a phase reversal but the reflection at an open boundary takes place without any phase change.' A phase reversal corresponds to a phase change of $\pi$ or $180^\circ$.

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