A string is stretched between fixed points separated by 75 . 0 cm . it is observed to have resonant frequency of 420 Hz and 315 Hz . There are no other resonant frequencies between these two. Then, the lowest resonant frequency for this string is

Step-by-step solution

For string fixed at both the ends, resonant frequency are given by v = n v 2 L Where symbols have their meaning. It is given that 315 Hz and 420 Hz are two consecutive resonant frequency, let these nth and (n+1)th harmonics. 315 = n v 2 L … i 420 = n + 1 v 2 L … i i ⟹Eq.(i)÷Eq.(ii) ⟹ 315 450 = n n + 1 ⟹ n = 3 From Eq. (i), lowest resonant frequency v 0 = v 2 L = 315 3 = 105 H z