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...
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.
The flat line represents the two waves adding together to get no sound.
Here, the two waves add together and generate a louder (larger amplitude) sound.
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.
`lambda/2=0.30 m, therefore lambda=0.60m`
`f=(343 m/s)/(0.60 m)=571.7 Hz~~570 Hz`
See Hyperphysics link.
p336 Physics by Giancoli 6th Edition