Q.A string of length $1\ \text{m}$, fixed at both ends, carries transverse waves of speed $200\ \text{m s}^{-1}$. Find the fundamental frequency at which the string can vibrate.

Stationary (standing) waves on a string arise from the superposition of an incident and a reflected travelling wave; the resulting resonant frequencies obey n = (1/2L)√(T/μ). The speed of a transverse wave on a stretched string is v = √(T/μ).

1. Formation of stationary waves and the laws of transverse waves on strings When a stretched string, fixed at both ends, is plucked or set vibrating, a transverse wave travels along it, reflects at the fixed end, and travels back. The forward-travelling wave $y_1 = a\sin(kx-\omega t)$ and the reflected wave $y_2 = a\sin(kx+\omega t)$ (reflection at a fixed end reverses the phase) superpose: $y = y_1 + y_2 = 2a\sin(kx)\cos(\omega t)$ This is a stationary wave: its amplitude $2a\sin(kx)$ depends only on position $x$ (not on time), so every point on the string oscillates in place with an amplitude fixed by its position — zero at fixed points called nodes ($\sin kx = 0$) and maximum at points called antinodes ($\sin kx = \pm1$), unlike a travelling wave whose disturbance pattern itself moves along the string. Because both ends of the string ($x=0$ and $x=L$) are fixed, they must be nodes; this boundary condition only allows wavelengths for which the string length $L$ is a whole number of half-wavelengths: $L = p\cdot\frac{\lambda}{2}, \quad p = 1,2,3,\dots$ Using $v = n\lambda$, where $v = \sqrt{T/\mu}$ is the wave speed on a string of tension $T$ and mass-per-unit-length $\mu$, the allowed (resonant) frequencies are: $n = \frac{p}{2L}\sqrt{\frac{T}{\mu}}, \quad p = 1,2,3,\dots$ These give the laws of transverse vibrations of a stretched string: for a given $\mu$, (i) frequency $n \propto 1/L$ (law of length), (ii) $n \propto \sqrt{T}$ (law of tension), and (iii) $n \propto 1/\sqrt{\mu}$ (law of mass).
2. Speed of the transverse wave on the given wire …