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A turning fork vibrating at 500 Hz fall...

A turning fork vibrating at `500 Hz` falls from rest accelerates at `10 m//s^(2)`.
How far below the point of release is the tuning fork when wave with a frequency of `475 Hz` reach the release point ?

A

`16.9 m`

B

`16 m`

C

`1.69 m`

D

`1.6 m`

Text Solution

Verified by Experts

The correct Answer is:
A
Distance travelled by the tuning fork
`= (1)/(2) "gt"^(2) = (1)/(2) xx 10 xx 1.837^(2) = 16.875 m`
`= 16.9 m`
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CENGAGE PHYSICS-SUPERPOSITION AND STANDING WAVES-Comprehension
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