Question

Two loudspeakers 6.0 m apart are playing the same frequency. If you stand 10.0 m in front of the plane of the speakers, centered between them, you hear a sound of maximum intensity. As you walk parallel to the plane of the speakers, staying 10.0 m in front of them, you first hear a minimum of sound intensity when you are directly in front of one of the speakers.

PART A

What is the frequency of the sound? Assume a sound speed of 340 m/s.

PART B

If you stay 10.0 m directly in front of one of the speakers, for what other frequencies between 100 Hz and 600 Hz is there a minimum sound intensity at this point?

Express your answer numerically. If there is more than one answer, enter your answers in ascending order separated by commas.

Answer 1

At the starting point centred between the speakers, as the sound is at a maximum, it means the speakers are in phase.

When standing exactly in front of one speaker, you ar 13 m from that speaker, and
root(6^2 + 13^2) = 14.3178 m from the other speaker.

For minimum sound, the speakers must be out 180 degrees out of phase.

So 14.3178 - 10 = 4.3178 m is an odd multiple of half wavelengths of the sound.
The wavelength of sound L * F = V

where L is the wavelenth in m,

F is the frequency in Hz, and V is the speed of sound in m/s.

So a half wavelength (H) = L/2
So
2H F / (2n + 1) = V
F = (2n + 1) * V / (2 * H) where n is an integer.

The lowest frequency is when n = 0

F = 340 / (2 * 4.3178) = 78.744 Hz

The next will be when n = 1
(2n + 1) = 3
F = 3 * 78.744 = 236.232 Hz
n = 2
F = 5 *78.744 = 393.72 Hz

F = 7 * 78.744 = 551.208 Hz [n=3]

The four frequencies between 100 and 800 Hz for minimum sound are
78.744, 236.232 , 393.72 and 551.208 Hz