Answer
(a) increases by a factor of $2$
(b) decreases by a factor of $3$
(c) increases by a factor of $2$
Work Step by Step
(a) We know that
$v=\sqrt{\frac{T}{\mu}}$
$\implies f=\frac{v}{\lambda}$
$\implies f=\frac{1}{2L}\sqrt{\frac{T}{\mu}}$
$\implies f=\frac{1}{2L}\sqrt{\frac{4T}{\pi D^2\rho}}$...eq(1) because $\mu=\pi \frac{D^2}{4}\rho$
If the tension in the string is increased by a factor of $4$ and the diameter and length remain the same, then from eq(1) frequency is a factor of $\sqrt 4$ and therefore the frequency increases by a factor of $2$.
(b) If the diameter of the string increases by a factor of $3$ while the tension $T$ and length $L$ remain the same, then from eq(1), we obtain:
$f=\frac{1}{2L}\sqrt{\frac{4T}{\pi(3D)^2}\rho}$
$\implies f=\frac{1}{2L}\sqrt{\frac{4T}{9\pi D^2\rho}}$
$\implies f=\frac{1}{3}(\frac{1}{2L}\sqrt{\frac{4T}{\pi D^2\rho}})$
Thus, the frequency decreases by a factor of $3$.
(c) If the length of the string is halved then
$f=\frac{1}{2(\frac{L}{2})}\sqrt{\frac{4T}{\pi D^2\rho}}$
$\implies f=2(\frac{1}{2L}\sqrt{\frac{4T}{\pi D^2\rho}})$
Thus, the frequency increases by a factor of $2$.
