Show that when a string fixed at its two ends vibrates in 1 loop, 2 loops, 3 loops and 4 loops, the frequencie

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Q. Show that when a string fixed at its two ends vibrates in 1 loop, 2 loops, 3 loops and 4 loops, the frequencies are in the ratio 1:2:3:4.

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Concept understanding — Stationary Waves on a String

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.

Note

A stationary wave is not a new kind of wave. It is the superposition of two identical travelling waves moving in opposite directions. No net energy flows past a node.

The Precise Statement

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

y_1 = A \sin(kx - \omega t)

y_2 = A \sin(kx + \omega t)

By the principle of superposition, the resultant displacement is:

y = y_1 + y_2 = A \sin(kx - \omega t) + A \sin(kx + \omega t)

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

y = 2A \sin(kx) \cos(\omega t)

This is the equation of a stationary wave.

y(x,t) = 2A \sin(kx) \cos(\omega t)

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. $x = (n+\frac12)\frac{\lambda}{2}$ .

Watch out

A common mistake is to think that the string is stationary everywhere. It is not — only the nodes are stationary. The rest of the string oscillates, but the pattern of nodes and antinodes does not move along the string.

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 $\lambda/4$ .

Feature	Condition	Position
Node	$\sin(kx)=0$	$x = n\frac{\lambda}{2}$
Antinode	$	\sin(kx)

Boundary Conditions: Why Only Certain Patterns Exist …

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A string fixed at both ends forms standing waves with nodes at the ends. The allowed wavelengths are $\lambda_n = 2L/n$ , and since frequency $f \propto 1/\lambda$ , the frequencies are $f_n = n f_1$ , giving the ratio $1:2:3:4$ for $n = 1,2,3,4$ .

Why This Works: The Physics of a Fixed String

When you pluck a string tied down at both ends, the ends cannot move — they are nodes (points of zero displacement). The only vibrations that survive are standing waves that "fit" perfectly between these two fixed points. This is called acoustic resonance: the string resonates only at specific frequencies, its natural frequencies or harmonics.

The key constraint: a standing wave on a string fixed at both ends must have a node at each end. Between them, you can have 1 loop (the fundamental), 2 loops (the first overtone), 3 loops, and so on. Each loop is half a wavelength.

L

The wave speed $v$ on the string is fixed by tension and linear density ( $v = \sqrt{T/\mu}$ ), so frequency $f$ and wavelength $\lambda$ are related by $v = f\lambda$ . Therefore:

f_n = \frac{v}{\lambda_n} = \frac{v}{2L/n} = n \cdot \frac{v}{2L}

Since $v$ and $L$ are constants for a given string, $f_n$ is directly proportional to $n$ .

Tip

The factor $\frac{v}{2L}$ is the fundamental frequency $f_1$ . So $f_n = n f_1$ — a beautifully simple result. The frequencies are just integer multiples of the lowest frequency.

Step-by-Step Verification

One loop ( $n=1$ ): The string vibrates in a single "belly". One loop = half a wavelength, so $\lambda_1 = 2L$ . Then $f_1 = v/(2L)$ .
Two loops ( $n=2$ ): Two loops fit into length $L$ , so $2(\lambda_2/2) = L$ → $\lambda_2 = L$ . Then $f_2 = v/L = 2(v/2L) = 2f_1$ . …

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Q.Show that when a string fixed at its two ends vibrates in 1 loop, 2 loops, 3 loops and 4 loops, the frequencies are in the ratio 1:2:3:4.

Why This Works: The Physics of a Fixed String

Step-by-Step Verification

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