Question

A string that is stretched between fixed supports separated by 87.1 cm has resonant frequencies of 798.0 and 684.0 Hz, with no intermediate resonant frequencies. What are (a) the lowest resonant frequency and (b) the wave speed?

Answer 1

Answer:

Lowest resonant frequency:  $F_0=114Hz$

Wave Speed: $v=198.36 \; m/s$

Explanation:

The lowest resonant frequency

For a fixed string a resonant mode is always  a integer multiple of fundamental frequency (the lowest resonant frequency):

$F_n =nF_0$

$F_n$ Is the frequency for a resonant mode

$F_0$ is fundamental frequency (the lowest resonant frequency).

Due to problem says  that there's not intermediate resonant frequencies between 684 Hz and 798 Hz, i.e, they are consecutive resonant modes, equations are as follows:

$684=nF_0 \; (1)\\798=(n+1)F_0 \; (2)$

Solving for n from equation (1) and replacing it in the equation (2)

$790=(\frac{684}{F_0} +1)F_0$

And finally solving for $F_0$

$789=684+F_0 =>F_0=114 \;Hz$

Wave speed

For a fixed string the wave speed can be calculated by  using

$F_0=\frac{V}{2L}$

Where V is the wave speed and L is the string length.

So, using this equation and solving for V:

$V=2*L*F_0 => V=2*(0.871 \; m)*114 \; Hz\\V=198.36 \;m/s$

It Is very important convert the string length unit to meters