Question

One end of a string of length $L$ is tied to the ceiling of a lift accelerating upwards with an acceleration $2g$. The other end of the string is free. The linear mass density of the string varies linearly from $0$ to $\lambda$ from bottom to top.

• A. The velocity of the wave in the string will be $0$.
• B. The acceleration of the wave on the string will be $3g/4$ every where.
• C. The time taken by a pulse to reach from bottom to top will be $\sqrt{8L/3g}.$
• D. The time taken by a pulse to reach from bottom to top will be $\sqrt{4L/3g}.$

Answer 1

Answer: (B), (C)

The acceleration of the wave on the string will be $3g/4$ every where.
The time taken by a pulse to reach from bottom to top will be $\sqrt{8L/3g}.$

The acceleration of string in ground frame will be $3g$. Now, draw fbd of piece of string that is $x$ units long from the bottom. Let tension at this distance be $T$.

$T-\left({\int }_0^x\frac{\lambda x}{L}dx\right)g=\left({\int }_0^x\frac{\lambda x}{L}dx\right)\left(2g\right)$

$T=\frac{3\lambda x^2}{2L}g$

Now, speed of wave in a string is,

$v=\sqrt{\frac{T}{\mu }}=\sqrt{\frac{\frac{3\lambda x^2}{2L}g}{\frac{\lambda x}{L}}}=\sqrt{\frac{3}{2}gx}$

Also, $a=\frac{dv}{dt}=v\frac{dv}{dx}=\frac{3g}{4}$

This is the acceleration of wave relative to the string.

Now, by $L=ut+\frac{1}{2}a{t}^2$, we get $t=\sqrt{\frac{8L}{3g}}$.