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. 105 Hz
B. 1.05 Hz
C. 1050 Hz
D. 10.5 Hz

Answer

A

Explanation

For string fixed at both the ends, resnant frequency are given by `f = (nv)/(2 L)`. It is given that `315 Hz` and `420 Hz` are two consecutive resonant frequencies. Let these be `nth` and `(n + 1) th` harmonices.
`315 = (nv)/(2 L)` ..(i)
`420 = ((n + 1)v)/(2 L)`...(ii)
Dividing Eq. (i) by Eq. (ii), we get
`(315)/(420) = (n)/( n + 1) rArr n = 3`
Lowest resonant frequency,
`f_0 = (v)/(2 L) = (315)/(3) = 105 Hz`.