First slide

Stationary waves

Question

The frequency of transverse vibrations in a stretched string is 200 Hz. If the tension is increased four times and the length is reduced to one-fourth the original value, the frequency of vibration will be

Easy

A

25 Hz
B

200 Hz
C

400 Hz
D

1600 Hz

Solution

$n = \frac{1}{2 l} \sqrt{\frac{T}{m}} \Rightarrow \frac{n_{1}}{n_{2}} = \frac{l_{2}}{l_{1}} \sqrt{\frac{T_{1}}{T_{2}}} = \frac{1}{4} \sqrt{\frac{1}{4}} = \frac{1}{8}$
$\Rightarrow n_{2} = 8 n_{1} = 8 \times 200 = 1600 Hz$

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