Question

The 2 speakers of a boom box are 35.0 cm apart. A single oscillator makes the speaker vibrate in phase at a frequency of 2.00kHz. At what angles, measured from the perpendicular bisector of the line joining the speakers, would a distant observer hear (a) maximum sound intensity?

(b) minimum sound intensity?

(Speed of sound = 340m/s)

Answer 1

Let's first find the wavelength of the wave from the wave speed equation:

$$v=f\lambda,$$ $$\lambda=\dfrac{v}{f}=\dfrac{340\ \dfrac{m}{s}}{2\cdot10^3\ Hz}=0.17\ m.$$

(a) Observer would hear the maximum sound intensity at the angles where there is a constructive interference of two sound waves:

$$dsin\theta=m\lambda.$$

Then, from this equation we can find $\theta$:

$$\theta=sin^{-1}(\dfrac{m\lambda}{d}).$$

Finally, we get:

$$\theta_0=sin^{-1}(\dfrac{0\cdot0.17\ m}{0.35\ m})=0^{\circ},$$ $$\theta_1=sin^{-1}(\dfrac{1\cdot0.17\ m}{0.35\ m})=29^{\circ},$$ $$\theta_2=sin^{-1}(\dfrac{2\cdot0.17\ m}{0.35\ m})=76.3^{\circ}.$$

For $m>2$ there are no solutions, since $sin\theta>1$.

(b) Observer would hear the minimum sound intensity at the angles where there is a destructive interference of two sound waves:

$$dsin\theta=(m+\dfrac{1}{2})\lambda.$$

Then, from this equation we can find $\theta$:

$$\theta=sin^{-1}(\dfrac{(m+\dfrac{1}{2})\lambda}{d}).$$

Finally, we get:

$$\theta_0=sin^{-1}(\dfrac{(0+\dfrac{1}{2})\cdot0.17\ m}{0.35\ m})=14^{\circ},$$ $$\theta_1=sin^{-1}(\dfrac{(1+\dfrac{1}{2})\cdot0.17\ m}{0.35\ m})=46.8^{\circ},$$

For $m>1$ there are no solutions, since $sin\theta>1$.