Answer:
a) [tex]f_1=5.587Hz[/tex]
b) [tex]f_{n+1}-f_n=5.587Hz[/tex]
Explanation:
The frequency of the [tex]n^{th}[/tex] harmonic of a vibrating string of length L, linear density [tex]\mu[/tex] under a tension T is given by the formula:
[tex]f_n=\frac{n}{2L} \sqrt{\frac{T}{\mu}[/tex]
a) So for the fundamental mode (n=1) we have, substituting our values:
[tex]f_1=\frac{1}{2(347m)} \sqrt{\frac{65.4\times10^6N}{4.35kg/m}}=5.587Hz[/tex]
b) The frequency difference between successive modes is the fundamental frequency, since:
[tex]f_{n+1}-f_n=\frac{n+1}{2L} \sqrt{\frac{T}{\mu}}-\frac{n}{2L} \sqrt{\frac{T}{\mu}}=(n+1-n)\frac{1}{2L} \sqrt{\frac{T}{\mu}}=\frac{n}{2L} \sqrt{\frac{T}{\mu}}=f_1=5.587Hz[/tex]