Q.Discuss the law of transverse vibrations in stretched strings.

Stationary Waves on a String

Imagine holding a skipping rope at one end while your friend holds the other. If you both flick your wrists at the same time, two identical pulses travel toward each other. When they meet, they pass through each other — but for an instant, the rope's shape is the sum of both pulses. Now imagine doing this continuously, sending wave after wave from both ends.

What happens when the timing is just right? The rope doesn't seem to travel anywhere. Instead, it vibrates in place, with some points completely still and others moving wildly. That's a stationary (or standing) wave.

The Intuition: Two Waves, One Pattern

A travelling wave on a string moves energy from one end to the other. But if you send a wave down the string and it reflects back from a fixed end, the original wave and the reflected wave are identical in amplitude and frequency but travel in opposite directions. Where they overlap, they interfere.

At certain points, the two waves always cancel each other out — those points never move. They are called nodes. At other points, the waves always add up constructively, giving maximum displacement — those are antinodes. The pattern is fixed in space; the wave does not travel. Hence the name: stationary wave.

The Precise Statement

Let two identical waves travel along a string in opposite directions. Their displacements are:

By the principle of superposition, the resultant displacement is:

Using the identity $\sin P + \sin Q = 2 \sin\frac{P+Q}{2} \cos\frac{P-Q}{2}$, we get:

This is the equation of a stationary wave.

What This Equation Tells You

The factor $\sin(kx)$ depends only on position $x$. The factor $\cos(\omega t)$ depends only on time $t$. This means:

• Every point on the string oscillates with the same frequency $\omega$, but with an amplitude that varies with position: $2A |\sin(kx)|$.
• Points where $\sin(kx) = 0$ are nodes — they never move. This happens when $kx = n\pi$, i.e. $x = n\frac{\lambda}{2}$, for $n = 0, 1, 2, \dots$
• Points where $|\sin(kx)| = 1$ are antinodes — they oscillate with maximum amplitude $2A$. This happens when $kx = (n+\frac12)\pi$, i.e. $x = (n+\frac12)\frac{\lambda}{2}$.

Nodes and Antinodes: The Fixed Pattern

The distance between two consecutive nodes (or two consecutive antinodes) is $\lambda/2$. The distance between a node and the next antinode is $\lambda/4$.

Boundary Conditions: Why Only Certain Patterns Exist …