A turning fork vibrating at `500 Hz` falls from rest accelerates at `10 m//s^(2)`.
Time taken by the waves with a frequency of `475 Hz` to reach the release point is nearly

Options

A. `1.79 s`
B. `1.84 s`
C. `17.9 s`
D. `18.4 s`

Answer

B

Explanation

Time taken `t' = (v_(s))/(g) = (17.9)/(10) = 1.79 s`
Time required to each the release point
`t = t'+ ((1/2 "gt"'^(2)))/(v)`
`= 1.79 + 1/2 xx (10 xx 1.79^(2))/(340)`
`= 1.79 + 0.047 = 1.837 s`