Waves: Common Mistakes and Fixes (3)

Question

A string of length $L = 1$ m and mass $m = 0.01$ kg is stretched with a tension $T = 100$ N. Find the frequency of the fundamental mode and the third harmonic. Both ends are fixed.

Solution — Step by Step

Why This Works

A string fixed at both ends must have a node at each end. The simplest standing wave fits exactly half a wavelength between the ends — that’s the fundamental. Higher harmonics fit $n$ half-wavelengths.

The wave speed depends only on the medium properties ($T$ and $\mu$), not on frequency. So once $v$ is fixed, frequency and wavelength are inversely related.

Alternative Method

Direct formula: $f_n = \frac{n}{2L}\sqrt{T/\mu}$. Plug in $n = 1, 3$:

$f_1 = \tfrac{1}{2}\sqrt{10000} = 50 \text{ Hz}, \quad f_3 = \tfrac{3}{2}\sqrt{10000} = 150 \text{ Hz}$

Common Mistake

The most expensive mistake on this kind of question is to mix up “third harmonic” with “third overtone.” For a both-ends-fixed string:

• Third harmonic = $n = 3$, so $f_3 = 3f_1$.
• Third overtone = the third frequency above the fundamental, which is also $n = 4$, so $f = 4f_1$.

When the question says “third harmonic,” it means $n = 3$. JEE 2023 had a question that used “third overtone” specifically to trip students up — read the wording.