Kvant Physics Problem 177
A stretched string of length $L$ carries transverse oscillations under constant tension $T$ and has linear mass density $\mu$, measured in $\mathrm{kg,m^{-1}}$.
Verified: no
Verdicts: SKIP + SKIP
Solve time: 55s
Source on kvant.digital
Problem
The problem was mistakenly published in the journal under number 167.
The frequency of oscillations of a string depends on its length, tension, and linear mass density—the mass per unit length of the string. Determine the form of these dependences.
I. A. Zaitsev
Setup and Assumptions
A stretched string of length $L$ carries transverse oscillations under constant tension $T$ and has linear mass density $\mu$, measured in $\mathrm{kg,m^{-1}}$. The oscillation frequency is denoted by $f$ with units $\mathrm{Hz}$. The string is assumed perfectly flexible and non-dissipative, with small transverse displacements so that linear wave theory applies. The tension is taken uniform along the string and independent of displacement. Endpoints are assumed fixed, producing standing waves.
The quantity to be determined is the functional dependence of $f$ on $L$, $T$, and $\mu$.
Physical Principles
Transverse waves on a stretched string satisfy the one-dimensional wave equation
$$\frac{\partial^2 y}{\partial t^2} = v^2 \frac{\partial^2 y}{\partial x^2},$$
where the wave speed is
$$v = \sqrt{\frac{T}{\mu}}.$$
Fixed boundary conditions at $x=0$ and $x=L$ require standing-wave solutions of the form
$$y(x,t) = A \sin(kx)\cos(\omega t),$$
with the condition
$$kL = n\pi, \quad n=1,2,3,\dots$$
The dispersion relation for nondispersive waves is
$$\omega = vk.$$
The oscillation frequency is related to angular frequency by
$$f = \frac{\omega}{2\pi}.$$
Derivation
The standing-wave condition $kL = n\pi$ gives
$$k = \frac{n\pi}{L}.$$
Substituting into the dispersion relation $\omega = vk$ yields
$$\omega = v \frac{n\pi}{L}.$$
Using $f = \frac{\omega}{2\pi}$ gives
$$f = \frac{1}{2\pi} v \frac{n\pi}{L} = \frac{n}{2L} v.$$
The wave speed follows from the balance between tension and inertia of a small string element. The transverse force from curvature produces acceleration proportional to $T$, while the inertia is set by $\mu$, giving
$$v = \sqrt{\frac{T}{\mu}}.$$
Substituting this into the frequency expression yields
$$f_n = \frac{n}{2L}\sqrt{\frac{T}{\mu}}.$$
The dependence on parameters is extracted by isolating the functional form
$$f \propto \frac{1}{L}\sqrt{\frac{T}{\mu}}.$$
Result
The oscillation frequency depends on length, tension, and linear mass density according to
$$f_n = \frac{n}{2L}\sqrt{\frac{T}{\mu}}.$$
No numerical data are provided in the statement, so the result remains in symbolic form:
$$\boxed{f_n = \frac{n}{2L}\sqrt{\frac{T}{\mu}} ;; \mathrm{Hz}}.$$
Sanity Checks
Dimensional consistency follows from $[T] = \mathrm{kg,m,s^{-2}}$ and $[\mu] = \mathrm{kg,m^{-1}}$, giving
$$\left[\sqrt{\frac{T}{\mu}}\right] = \sqrt{\frac{\mathrm{kg,m,s^{-2}}}{\mathrm{kg,m^{-1}}}} = \sqrt{\mathrm{m^2,s^{-2}}} = \mathrm{m,s^{-1}}.$$
Dividing by $L$ produces units $\mathrm{s^{-1}}$, consistent with frequency.
Increasing tension increases wave speed and therefore frequency through the factor $\sqrt{T}$, while increasing linear density increases inertia and reduces frequency through $\sqrt{1/\mu}$. Doubling the length reduces the frequency by a factor of two due to the inverse proportionality in the standing-wave condition. The dependence remains consistent for all harmonics through the factor $n$, which counts nodes along the string.