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. `105Hz`
B. `155Hz`
C. `205Hz`
D. `315Hz`

Answer

A

Explanation

For a string stretched between two fixed points, the two successive resonant frequencies for a string will be
`(n upsilon)/(2l)` and `((n+1)upsilon)/(2l)`.
Given, `((n+1)upsilon)/(2l)=420Hz` and `(n upsilon)/(2l)=315Hz.`
`:. ((n+1)upsilon)/(2l)-(n upsilon)/(2l)=420-315=105`
or `(upsilon)/(2l)=105Hz`