A string fixed at both the ends is vibrating in the lowest mode of vibration for which a point at quarter of its length from one end is a point of maximum amplitude. The frequency of vibration in this mode is 100 Hz. What will be the frequency emitted when it vibrates in the next mode such that the point is again a point of maximum amplitude ?

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300 Hz

600 Hz

400 Hz

200 Hz

answer is D.

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Detailed Solution

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Point P is an antinode i,e the string is vibrating in its second harmonic. Let $f_{0}$ be the fundamental frequency. Then
$2 f_{0} = 100 H z$ $∴ f_{0} = 50 H z$
Now P is an antinode (at length $l / 4$ from one end) so centre should be a node. So, next higher frequency will be sixth harmonic or $6 f_{0}$ which is equal to 300 Hz as shown below.
Question Image

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