A string is stretched between fixed points separated by `75.0cm`. It is observed to have resonant frequencies of `420 Hz` and `315 Hz`. There are no other resonant frequencies between these two. Then, the lowest resonant frequency for this string is

Options

A. (a) `105 Hz`
B. (b) `1.05 Hz`
C. ( c ) `1050 Hz`
D. (d) `10.5 Hz`

Answer

A

Explanation

(a) Given `(nv)/(2l) = 315` and `(n + 1)(v)/(2l) = 420`
rArr `(n + 1)/(n) = (420)/(315)` rArr `n = 3`
Hence `3 xx (v)/(2l) = 315` rArr `(v)/(2l) = 105 Hz`
lowest resonant frequency is when `n = 1`
Therefore lowest resonant frequency `= 105 Hz`.