Problem 13.35

13.35.
The maximum speed of the speaker occurs at the equillibrium location, when all the energy in the spring connected to the speaker is in kinetic energy. There,
$\frac{1}{2} m v_{max}^{2} = \frac{1}{2} k A^{2}$
Solving for v_max, we find
$v_{max} = \sqrt{\frac{k}{m}} A$
So, the frequencies heard range from
$f'_{min} = f(\frac{v}{v + \sqrt{\frac{k}{m}}A})$
to
$f'_{max} = f(\frac{v}{v - \sqrt{\frac{k}{m}}A})$
where v is the speed of sound.
Plugging in the numbers, we find that
f'_min = (440 Hz)(.997)
f'_min = 439 Hz
f'_max = (440 Hz)(1.003)
f'_max = 441 Hz