Waves: Application Problems (7)

easy • 3 min read
Tags Waves

Question

A string of length 1 m1 \text{ m} and mass 4 g4 \text{ g} is stretched with a tension of 100 N100 \text{ N}. Find (a) the speed of a transverse wave on the string, (b) the fundamental frequency when both ends are fixed, and (c) the frequency of the third harmonic.

Solution — Step by Step

Linear mass density:

μ=mL=4×10−31=4×10−3 kg/m\mu = \frac{m}{L} = \frac{4 \times 10^{-3}}{1} = 4 \times 10^{-3} \text{ kg/m}

Wave speed:

v=Tμ=1004×10−3=25000=158.1 m/sv = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{100}{4 \times 10^{-3}}} = \sqrt{25000} = 158.1 \text{ m/s}

For a string fixed at both ends, the fundamental wavelength is λ1=2L=2 m\lambda_1 = 2L = 2 \text{ m}.

f1=vλ1=158.12≈79.1 Hzf_1 = \frac{v}{\lambda_1} = \frac{158.1}{2} \approx 79.1 \text{ Hz}

Allowed harmonics are integer multiples: fn=nf1f_n = n f_1. So the third harmonic:

f3=3f1≈237.2 Hzf_3 = 3 f_1 \approx 237.2 \text{ Hz}

Why This Works

The wave speed on a string depends only on the medium (tension and mass per unit length), not on how you shake it. The frequency is set by the boundary conditions — fixed ends force a node at each end, which selects integer multiples of the fundamental.

That’s why a guitarist tunes by changing tension (changes vv, changes all fnf_n) and frets by changing length (changes λ\lambda, changes all fnf_n). Both knobs work because f=v/λf = v/\lambda.

Alternative Method

You can write the result in one formula:

fn=n2LTμ,n=1,2,3,…f_n = \frac{n}{2L}\sqrt{\frac{T}{\mu}}, \quad n = 1, 2, 3, \ldots

Plug in directly: f3=32×11000.004=1.5×158.1≈237.2 Hzf_3 = \dfrac{3}{2 \times 1}\sqrt{\dfrac{100}{0.004}} = 1.5 \times 158.1 \approx 237.2 \text{ Hz}.

For a pipe open at both ends, the formula is the same as a fixed-fixed string. For a pipe closed at one end, only odd harmonics are allowed: fn=(2n−1)v/(4L)f_n = (2n-1)v/(4L). NEET loves to flip between these three cases, so memorise all three boundary conditions.

Common Mistake

A frequent slip is using λ1=L\lambda_1 = L instead of 2L2L. The fundamental for a fixed-fixed string fits half a wavelength in the length LL (one antinode in the middle, nodes at the ends). So L=λ/2L = \lambda/2, giving λ1=2L\lambda_1 = 2L. Drawing the standing-wave shape before plugging numbers prevents this.

The other slip is unit confusion on μ\mu — forgetting to convert 4 g4 \text{ g} to 0.004 kg0.004 \text{ kg} multiplies your wave speed by 1000\sqrt{1000}, an absurd answer that should immediately fail your sanity check.

Final answer: v≈158.1 m/sv \approx 158.1 \text{ m/s}, f1≈79.1 Hzf_1 \approx 79.1 \text{ Hz}, f3≈237.2 Hzf_3 \approx 237.2 \text{ Hz}.

Want to master this topic?

Read the complete guide with more examples and exam tips.

Go to full topic guide →

Try These Next

A Wave Has Frequency 500 Hz and Wavelength 0.6 m — Find Speed

easy

Beats phenomenon — how two waves of slightly different frequency create beats

easy

Doppler effect — all four cases for sound, formula selection

medium

Doppler Effect: Ambulance Moving Towards You at 30 m/s — Apparent Frequency

hard

Doppler effect — all four cases for sound, formula selection

medium