Answer
(a) $13.8Hz$
(b) increase
(c) $18.3Hz$
Work Step by Step
(a) We can find the required beat frequency as follows:
$L=\frac{v}{2f_1}$
$L=\frac{34.2}{2(212)}=8.066cm$
and $L^{\prime}=L+\Delta L$
$L^{\prime}=8.066+0.560=8.626cm$
Now $f_{beat}=|f_1-f_1^{\prime}|=|\frac{v}{2L}-\frac{v}{2L^{\prime}}|$
We plug in the known values to obtain:
$f_{beat}=|\frac{34.2}{2(8.066\times 10^{-2})}-\frac{34.2}{2(8.626\times 10^{-2})}|=13.8Hz$
(b) We know that the fundamental frequency is decreased and the beat frequency will increase if the length of the longer string is increased.
(c) $f_{beat}=|f_1-f_1^{\prime}|=|\frac{v}{2L}-\frac{v}{2L^{\prime}}|$
We plug in the known values to obtain:
$f_{beat}=|\frac{34.2}{2(8.066\times 10^{-2})}-\frac{34.2}{2(8.066\times 10^{-2}+0.761)\times 10^{-2}}|=18.3Hz$
