A horizontal string of linear density 0.02 kg/m is stretched with a tension of 120 N. A transverse sinusoidal disturbance of amplitude 5 mm and angular frequency 80 rad/s travels along the string. Calculate (a) the speed of the wave, (b) the wavelength, and (c) the maximum transverse force exerted by the wave at a point on the string.

Step-by-step solution

1. The speed of a transverse wave on a string is $v=\sqrt{T/mu}$.
2. Insert $T=120\mathrm{N}$ and $mu=0.02kg/m$ to get $v=\sqrt{120/0.02}=\sqrt{6000}approx77.46m/s$. Round to $77.5m/s$ for part (a).
3. Frequency $f$ is related to angular frequency by $f=\omega /(2\pi )=80/(2\pi )\approx 12.73\mathrm{Hz}$. Use $v=flambda$ to find $\lambda =\frac{77.46}{12.73}\approx 6.09\mathrm{m}$ for part (b).
4. For a sinusoidal wave $y=Asin(kx-\omega t)$ the maximum transverse force per unit length is $T\frac{d\theta }{dx}$. The maximum slope $d\theta /dx$ occurs when $cos(kx-\omega t)=pm1$, giving $|d\theta /dx{|}_{max}=Ak$. Compute $k=2\pi /\lambda =2\pi /6.09\approx 1.033rad/m$. Thus $F_{max}=TcdotAcdotk=120\times 0.005\times 1.033\approx 0.619\mathrm{N}$, rounded to $0.62\mathrm{N}$ for part (c).

Concept tested: Wave properties on a string

Published 13 October 2026 · How this solution was prepared and checked