A string is stretched between two rigid supports separated by 75 ~cm . There are no resonant frequencies between 420 ~Hz and 315 ~Hz . The lowest resonant frequency for the string is

Step-by-step solution

As there is no resonant frequency between 315 ~Hz and 420 ~Hz , let 315 ~Hz be n ^ th overtone and 420 ~Hz be ( n +1)^ th overtone. Now, v= nv 2 l 315= nv 2 l and 420= ( n +1) v 2 l Taking the ratio, array ll & 315 420 = n n+1 & 315 n+315=420 n & n=3 array The resonant frequency is v₀= v 2 l Therefore, from equation (i) we get, v₀= v n = 315 3 =105 ~Hz