Question

Figure shows an air filled, acoustic interfero meter, used to demonstrate the interference of sound waves. Sound source $S$ is an oscillating diaphragm; $D$ is a sound detector, such as the ear or a microphone. Path $SBD$ can be varied in length, but path $SAD$ is fixed. At $D,$ the sound wave coming along path $SBD$ interferes with that coming along path $SAD.$ In one demonstration, the sound intensity at $D$ has a minimum value of 100 units at one position of the movable arm and continuously climbs to a maximum value of 900 units when that arm is shifted by $1.65cm.$ Find the frequency of the sound emitted by the source.

Answer 1

When the right side of the instrument is pulled out a distance $d$, the path length for sound waves increases by $2d$.
since the interference pattern changes from a minimum to the next maximum, this distance must be half a wavelength of the sound.
So $2d=\lambda /2$ where $\lambda$ is the wavelength.
Thus $\lambda =4d$ and, if $v$ is the speed of sound, the frequency is
$f=v/\lambda =v/4d=\left(343m/s\right)/4\left(0.0165m\right)=5.2\times {10}^{3}Hz$