Animated Solution for Physics - Waves: A uniform string of length 20 m is suspended from a rigid support. A short wave pulse is introduced at its lowest end. It starts moving up the string. The time taken to reach the support is (Take, $g=10{\text{ms}}^{-2}$)

This problem is a beautiful intersection of wave mechanics and calculus. It forces us to abandon the comfort of constant velocity and deal with a dynamically changing system. Let's break down the physics of a wave pulse traveling up a heavy, hanging string.

When a string is massless, the tension is uniform throughout. But here, the string has a significant mass. Imagine isolating a small segment of the string at a height $x$ from the bottom. What is pulling this segment down? It's the weight of the string below it.

If the string has a total mass $m$ and length $L$, its linear mass density is $\mu =\frac{m}{L}$. The mass of the portion of the string below our segment is simply $\mu x$. Therefore, the tension $T$ at a height $x$ is equal to the weight of this lower portion:

Notice that at the very bottom ($x=0$), the tension is zero. At the very top ($x=L$), the tension is maximum, equal to the total weight of the string, $mg$.

The speed of a transverse wave on a string is governed by the classic formula $v=\sqrt{\frac{T}{\mu }}$. Because our tension $T$ is a function of $x$, our wave speed $v$ will also be a function of $x$. Let's substitute our expression for tension:

This is a fascinating result! The mass density $\mu$ completely cancels out. The speed of the wave pulse depends only on the acceleration due to gravity $g$ and its current height $x$ from the bottom. As the pulse climbs higher, it accelerates.

Integration

To find the total time taken, we must acknowledge that velocity is the rate of change of position, $v=\frac{dx}{dt}$. We can set up a differential equation:

Separating the variables to group $x$ terms with $dx$ and $t$ terms with $dt$:

Now, we integrate both sides. The time $t$ goes from $0$ to the total time $T_{total}$, while the position $x$ goes from the bottom ($0$) to the top ($L$):

This is a standard, highly useful result to remember for JEE. Substituting the given values $L=20\text{ m}$ and $g=10{\text{ m/s}}^2$:

Constant Acceleration

While integration is the rigorous way, there is a brilliant shortcut hidden in the kinematics. Let's look at the acceleration of the wave pulse. We know $v=\sqrt{gx}$. Using the chain rule, acceleration $a=\frac{dv}{dt}=v\frac{dv}{dx}$:

The acceleration of the wave pulse is constant! It is exactly half of the acceleration due to gravity. Since the acceleration is constant, we can bypass integration entirely and use the standard equations of motion. Using $S=ut+\frac{1}{2}a{t}^2$ with initial velocity $u=0$ (since $v=\sqrt{g(0)}=0$ at the bottom):

We arrive at the exact same formula in seconds! This dual perspective—seeing the problem through both calculus and pure kinematics—is the hallmark of a true physics master.