Two identical loudspeakers are some distance apart. A person stands 5.70 m from one speaker and 2.80 m from the other. What is the fourth lowest frequency at which destructive interference will occur at this point? The speed of sound in air is 343m/s.
Destructive interference of two identical soundwaves occurs, if the sound waves are exactly shifted by one half of their wavelength. That means the difference in path length from source to observer need to be an integer multiple of half od wavelength:
n·(lamda/2) = dL (with n=1,2,3.....
So destructive interference occurs, if the wavelength is:
lamda = 2·dL/n
Corresponding frequencies are
f = c/lamda = c·n / (2·dL)
The fourth frequency is for n=4,
f_min = 4 *c/(2·dL) = 343m/s * 4 / ( 2· (5.7m - 2.8 m)) = 236.55 Hz
the frequency is 236.55 Hz
Get Answers For Free
Most questions answered within 1 hours.