570 Hz (± 3 Hz)

**Assumptions** Based on your description, I assumed that the speakers are both on the x-axis. Speed of sound in air at 20° C is 343 m/s (substitute another value based on what is commonly used - hence the "±") Another assumption is that we did...

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570 Hz (± 3 Hz)

**Assumptions** Based on your description, I assumed that the speakers are both on the x-axis. Speed of sound in air at 20° C is 343 m/s (substitute another value based on what is commonly used - hence the "±") Another assumption is that we did not miss any other nodes.

**Background** Interference occurs due to superposition of waves; the energy carried by the two waves can occupy the same place at the same time. Waves can be visualized as a sine or cosine wave that has peaks and valleys. When you superimpose two waves over each other, you end up with interference. When the frequency is the same, then you get perfect constructive or destructive interference as in the problem. If the peaks and valleys of these two waves line up you get constructive interference, giving the effect of louder sound. When the peaks line up with valleys, you get destructive interference and you end up with the effect of no sound at all. Out of phase basically means that if you were to superimpose the two waves, a peak would line up with a valley.

**Approach**

Draw it!

Notice that by moving the out-of-phase waves by 1/2 wavelength, they line up either constructively or destructively. This means that 30 cm is 1/2 wavelength. Thus we are on our way to solving.

`f=v/lambda` , where f is the frequency in Hz, v is the speed of sound, and `lambda`

is the wavelength in meters.

**Calculate**

`lambda/2=0.30 m, therefore lambda=0.60m`

`f=(343 m/s)/(0.60 m)=571.7 Hz~~570 Hz`

**References**

See Hyperphysics link.

p336 Physics by Giancoli 6th Edition

**Further Reading**

The two speakers produce the loudest sound when the waves of both speakers add with constructive interference. Meaning, when the speakers are 30 cm apart, they are adding the waves thus making the sound at its loudest. However when the distance is increasing, the destructive interference starts to dominate. Consequently it decreases the sound. At 60 cm apart, the two waves “cancel” each other thus no sound will be produced. Now you can notice that as you will increase the distance again, the sound will increase again. This is because in the waveform, the distance between the peak and the trough is 1/2 of the wavelength. 30cm will now be the 1/2 of the wavelength and 60cm must be the full wavelength.

To solve for the frequency of the sound, we know that:

`v = lambda * f`

Therefore,

`f = (v)/(lambda)`

v= speed of sound = 343 m/s (in air, 20^o Celsius at sea level)**

`lambda = 60 cm * (1m)/(100cm) ` = 0.6 m

`f = (343 m/s)/(0.60m)`

f = **571.67 s^-1**` ` = frequency

**speed of sound varies