Kvant Physics Problem 177

Verified: yes
Verdicts: PASS + PASS
Solve time: 6m51s
Source on kvant.digital

Problem

The problem was erroneously published in the journal under number 167.

The frequency of vibration of a string depends on its length, tension, and linear density, that is, the mass per unit length of the string. Determine the form of these dependences.

I. A. Zaitsev

Setup and Assumptions

A stretched string performs small transverse vibrations. The quantity to be determined is the vibration frequency $f$ of the string.

The frequency is assumed to depend only on the string length $l$, the tension force $T$, and the linear density $\mu$, where

$$\mu=\frac{m}{l}.$$

The dimensions and units of the quantities are

$$[f]=T^{-1},$$

$$[l]=L,$$

$$[T]=MLT^{-2},$$

$$[\mu]=ML^{-1}.$$

The string is treated as ideal. Its bending stiffness is neglected, the tension is uniform along the string, the oscillation amplitudes are small, and the surrounding medium does not influence the motion. Since no numerical values are given, the task is to determine the functional dependence of the frequency on the parameters by dimensional analysis.

Physical Principles

The solution is based on the principle of dimensional homogeneity.

Any physically meaningful equation must have identical dimensions on both sides. If the frequency depends only on the quantities $l$, $T$, and $\mu$, then it must be expressible in the form

$$f=C,l^{a}T^{b}\mu^{c},$$

where $C$ is a dimensionless numerical constant and the exponents $a$, $b$, and $c$ are to be determined from dimensional consistency.

Derivation

Assume

$$f=C,l^{a}T^{b}\mu^{c}.$$

Substituting the dimensions of all quantities gives

$$T^{-1} = L^{a} \left(MLT^{-2}\right)^{b} \left(ML^{-1}\right)^{c}.$$

Collecting powers of the fundamental dimensions $M$, $L$, and $T$ yields

$$T^{-1} = M^{,b+c} L^{,a+b-c} T^{-2b}.$$

The exponents of corresponding dimensions must be equal on both sides.

For the dimension of mass,

$$b+c=0.$$

For the dimension of length,

$$a+b-c=0.$$

For the dimension of time,

$$-2b=-1.$$

From the last equation,

$$b=\frac12.$$

Substituting into the first equation gives

$$c=-\frac12.$$

Substituting these values into the second equation gives

$$a+\frac12+\frac12=0,$$

hence

$$a=-1.$$

The dependence of the frequency is therefore

$$f=C,l^{-1}T^{1/2}\mu^{-1/2}.$$

Rearranging,

$$f=C,\frac{1}{l}\sqrt{\frac{T}{\mu}}.$$

Dimensional analysis determines the form of the dependence but not the dimensionless constant $C$.

For a string fixed at both ends and vibrating in its fundamental mode, wave theory gives

$$C=\frac12,$$

so that

$$f=\frac{1}{2l}\sqrt{\frac{T}{\mu}}.$$

Result

The required dependence of the vibration frequency on length, tension, and linear density is

$$\boxed{,f\propto \frac{1}{l}\sqrt{\frac{T}{\mu}}, }.$$

More explicitly,

$$\boxed{,f=C,\frac{1}{l}\sqrt{\frac{T}{\mu}},},$$

where $C$ is a dimensionless constant determined by the vibration mode and boundary conditions.

For the fundamental vibration of a string fixed at both ends,

$$\boxed{,f=\frac{1}{2l}\sqrt{\frac{T}{\mu}},}.$$

No numerical substitution can be performed because the problem provides no numerical values.

Sanity Checks

The dimensions of the expression are

$$\left[\frac{1}{l}\sqrt{\frac{T}{\mu}}\right] = L^{-1} \sqrt{\frac{MLT^{-2}}{ML^{-1}}} = L^{-1}\sqrt{L^2T^{-2}} = T^{-1},$$

which coincides with the dimension of frequency.

If the tension $T$ increases, the frequency increases as $\sqrt{T}$. A more strongly stretched string vibrates faster, which agrees with experience.

If the linear density $\mu$ increases, the frequency decreases as $1/\sqrt{\mu}$. A heavier string vibrates more slowly, which is physically reasonable.

If the length $l$ increases, the frequency decreases in inverse proportion to $l$. Longer strings produce lower pitches, matching the behavior of musical instruments.

The step most susceptible to error is the solution of the system of exponent equations. A sign mistake in the relation

$$a+b-c=0$$

or in the dimension of $\mu$, namely

$$[\mu]=ML^{-1},$$

would lead to an incorrect power of the length and spoil the final result.