A string is stretched between fixed points separated by 75 . 0 cm. It is observed to have resonant frequencies of 42O Hz and. 315 Hz. There are no other resonant frequencies between these two. Then, the lowest resonant frequency for this string is

The resonant frequency for a string fixed at both the ends is given byf=nV2LIt is given that 315 Hz and 420 Hz are two consecutive resonant frequencies. Let the two frequencies are corresponding to nth and (n+1)th harmonics. We have 315=nv2L Dividing eq. (1) by eq. (2), we Bet 315420=n(n+1)  or  n=3 So, the lowest resonant frequency f0=V2L=315n=3153=105Hz