Question

A string that is stretched between fixed supports separated by 92.3 cm has resonant frequencies of 1140 and 1026 Hz, with no intermediate resonant frequencies. What are (a) the lowest resonant frequency and (b) the wave speed?

Answer 1

a) The resonant wavelengths have the general formula ( 2 . L / n )

So we put wavelength 1 ( call it w1 ) = 2L / n and w2 will be 2L /(n+1)

F1 = c / w1 and F2 = c / w2

then

F1w1 = F2w2

1026 . 153/n = 1140 . 153 / (n+1)

the 153 cancels

1140( n + 1) = 1026 . n

n = 1140 /( 1140-1026) = 10

Therefore 1140 is the 10th harmonic
The first harmonic ( the fundamental, the lowest resonant frequency) must be 1140 / 10 = 114 Hz

b) wavespeed = F . w = 114 . 153 = 17442 = 17442 m/s