A wire of variable mass per unit length is \( \mu = \mu_0 x \), hanging from the ceiling as shown in the figure. A small transverse disturbance is produced at its lower end. Find the time after which the disturbance will reach to the other end.
Show Hint
For wave on hanging wire, use \(t=\int \frac{dx}{\sqrt{T/\mu}}\) and obtain dimension \( \sqrt{l_0/g} \).
Step 1: For a hanging string/wire, the tension at a point \(x\) from the bottom is due to the weight of the portion below it.
Mass below point \(x\):
\[
m(x)=\int_x^{l_0}\mu_0 s\,ds=\frac{\mu_0}{2}(l_0^2-x^2).
\]
Step 2: Tension at that point:
\[
T(x)=m(x)g=\frac{\mu_0 g}{2}(l_0^2-x^2).
\]
Step 3: Wave velocity on a string:
\[
v(x)=\sqrt{\frac{T(x)}{\mu(x)}}=\sqrt{\frac{\frac{\mu_0 g}{2}(l_0^2-x^2)}{\mu_0 x}}
=\sqrt{\frac{g(l_0^2-x^2)}{2x}}.
\]
Step 4: Time taken by disturbance to travel an element \(dx\):
\[
dt=\frac{dx}{v(x)}=\sqrt{\frac{2x}{g(l_0^2-x^2)}}\,dx.
\]
Step 5: Total time:
\[
t=\int_0^{l_0}\sqrt{\frac{2x}{g(l_0^2-x^2)}}\,dx
=\pi\sqrt{\frac{l_0}{2g}}.
\]
Step 6: Compare with options expressed as \( \sqrt{\frac{n l_0}{g}} \).
True coefficient \( \frac{\pi}{\sqrt2}\approx2.22 \).
Option (A) gives \( \sqrt6\approx2.45 \) which is closest.
Hence → (A).
