Question

A loudspeaker having a diaphragm that vibrates at 910 Hz is traveling at 75.0 m/s directly toward a pair of holes in a very large wall in a region for which the speed of sound is 344 m/s. You observe that the sound coming through the openings first cancels at 12.9 ∘ with respect to the original direction of the speaker when observed far from the wall.(a) How far apart are the two openings? (b) At what angles would the sound first cancel if the source stopped moving?

Answer 1

To solve this problem we need to apply doppler equation,

Our values are:

$f_0 = 910Hz\\c = 75m\\v=344m/s\\\theta= 12.9\°$

Doppler equation is given by,

$f= \frac{v}{v-c} f_0$

Substituting,

$f= \frac{344}{344-75}*910$

$f = 1163Hz$

$\lambda = \frac{344}{1163}$

Wavelenght is equal to,

$\lambda = \frac{c}{f}$

$\lambda = 0.2956$

We can now find the distance by,

$d= \frac{1}{2}\frac{\lambda}{sin\theta}$

$d=\frac{1}{2}\frac{0.2956}{sin(12.9)}$

$d= 0.66m$

b) We can find at what angles would sound first cancel through the same equation of wavelenght,

$\lambda = \lambda{c}{f} = \frac{344}{1163}$

$d* sin\theta = \frac{1}{2}*\lambda$

$\theta = arcsin( 0.5*(344/1163)/0.66)=12.94\°$