- #1

- 53

- 5

## Homework Statement

A mass [itex]m[/itex] attached to a spring of spring constant [itex]k[/itex] emits sound at frequency [itex]f[/itex], detected by a collinear observer at distance [itex]r[/itex]. If the mass has maximum velocity [itex]v_0[/itex], what is the total number of waves the observer detects in one period of oscillation?

## Homework Equations

Unless I'm missing something, we should only need the Doppler effect for a moving source:

[tex]f'=\frac{v_s}{v_s-v_0}f[/tex]

the angular velocity of a spring-mass system

[tex]\omega^2=\frac{k}{m}[/tex]

and the number of waves being the integral of the frequency over time

[tex]N=\int_{t_i}^{t_f}f\,\mathrm{d}t[/tex]

This last one I'm not sure about, but it makes sense in the case that [itex]f[/itex] is constant, and seems a natural generalization.

## The Attempt at a Solution

The spring-mass system oscillates with velocity given by

[tex]v(t)=v_0\cos\omega t,\qquad\textrm{where}\qquad\omega^2=\frac{k}{m}.[/tex]

The observer is stationary, and so detects a Doppler-shifted frequency [itex]\tilde{f}[/itex] given by

[tex]\tilde{f}(t)=\frac{v_s}{v_s-v(t)}f=\frac{v_s}{v_s-v_0\cos\omega t}.[/tex]

The number of waves detected by the observer is simply the integral of frequency over time:

[tex]N=\int_0^T\tilde{f}(t)\,\mathrm{d}t=f\int_{-\frac{\pi}{\omega}}^\frac{\pi}{\omega}\frac{v_s\,\mathrm{d}t}{v_s-v_0\cos\omega t}.[/tex]

We are assuming [itex]v_0<v_s[/itex], so that the denominator is never zero. We first substitute [itex]\theta=\omega t[/itex], so [itex]\mathrm{d}t=\mathrm{d}\theta/\omega[/itex] and [itex]\theta(\pm\frac{\pi}{\omega})=\pm\pi[/itex]:

[tex]N=\frac{v_sf}{\omega}\int_{-\pi}^\pi\frac{\mathrm{d}\theta}{v_s-v_0\cos\theta}.[/tex]

We now substitute [itex]\theta=2\phi[/itex], so [itex]\mathrm{d}\theta=2\,\mathrm{d}\phi[/itex] and [itex]\phi(\pm\pi)=\pm\frac{\pi}{2}[/itex]:

[tex]N=\frac{v_sf}{\omega}\int_{-\frac{\pi}{2}}^\frac{\pi}{2}\frac{2\,\mathrm{d}\phi}{v_s-v_0\cos 2\phi}

=\frac{2v_sf}{\omega}\int_{-\frac{\pi}{2}}^\frac{\pi}{2}\frac{\mathrm{d}\phi}{v_s(\cos^2\phi+\sin^2\phi)-v_0(\cos^2\phi-\sin^2\phi)}[/tex]

[tex]=\frac{2v_sf}{\omega}\int_{-\frac{\pi}{2}}^\frac{\pi}{2}\frac{\sec^2\phi\,\mathrm{d}\phi}{\sec^2\phi[(v_s-v_0)\cos^2\phi+(v_s+v_0)\sin^2\phi]}

=\frac{2v_sf}{\omega}\int_{-\frac{\pi}{2}}^\frac{\pi}{2}\frac{\sec^2\phi\,\mathrm{d}\phi}{(v_s-v_0)+(v_s+v_0)\tan^2\phi}.[/tex]

We now substitute [itex]x=\tan\phi[/itex], so [itex]\mathrm{d}x=\sec^2\phi\,\mathrm{d}\phi[/itex], and [itex]x(\pm\frac{\pi}{2})=\pm\infty[/itex]:

[tex]N=\frac{2v_sf}{\omega}\int_{-\infty}^\infty\frac{\mathrm{d}x}{(v_s-v_0)+(v_s+v_0)x^2}.[/tex]

We finally substitute [itex]x=\sqrt{(v_s-v_0)/(v_s+v_0)}z[/itex], so [itex]\mathrm{d}x=\sqrt{(v_s-v_0)/(v_s+v_0)}\,\mathrm{d}z[/itex]:

[tex]N=\frac{2v_sf}{\omega}\sqrt\frac{v_s-v_0}{v_s+v_0}\int_{-\infty}^\infty\frac{\mathrm{d}z}{(v_s-v_0)+(v_s-v_0)z^2}

=\frac{2v_sf}{\omega}\frac{1}{\sqrt{v_s^2-v_0^2}}\int_{-\infty}^\infty\frac{\mathrm{d}z}{1+z^2}

=\fbox{$\displaystyle\frac{2\pi fv_s}{\sqrt{v_s^2-v_0^2}}\sqrt\frac{m}{k}$.}[/tex]

There is also the "obvious" answer

[tex]n=2\pi f\sqrt\frac{m}{k}.[/tex]

I can see why this answer should be right, and why qualitatively my answer above should be wrong. However, I cannot find any flaw in my quantitative argument. Can anybody help me spot the error in my reasoning?

Last edited: