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

Options

A. 10.5 Hz
B. 105 Hz
C. 155 Hz
D. 205 Hz

Explanation

For a string fixed at both ends, the resonant frequencies are
`upsilon_(n)=(nv)/(2L)` where `n=1,2,3,`….
The difference between two consecutive resonant frequencies is
`Deltaupsilon_(n)=upsilon_(n+1)-upsilon_(n)=((n+1)v)/(2L)-(nv)/(2L)=(v)/(2L)`
Which is also the lowest resonant frequency `(n=1`).
Thus the lowest resonant frequency for the given string.
`=420Hz-315Hz=105Hz`