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Q.

A wire of length 0.4m stretched at both ends vibrates 250 times per second. If the length of the wire is increased by 0.1m and the stretching force is reduced to (14)th of its original value then the new frequency is

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a

150Hz

b

50Hz

c

100Hz

d

75Hz

answer is C.

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Detailed Solution

∴f=12LTμ

250=12×0.4Tμ→1

∴f1=12×0.5T4μ→2(∴T1=T4)

Eq(1)Eq(2)⇒250f1=54×2

f1=100Hz

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