course phys 202
1. Beat frequency at 262 and 277 Hz; beat frequency two octaves lower-277-262= 15Hz. Depending on the octave that determines how much it is reduced, in this case it is by two so it would be .25 * 15Hz= 3.75Hz
2. speakers 1.8 meters apart, listener three meters from one and 3.5 m from the other **** gen phy what is the lowest frequency that will permit destructive interference at the location of the listener?
- dL= lamda/2
= 2(3.5-3.00
= 1m
f=v/lamda
= 343/1
= 343Hz
'dL=3'lambda/2
=2/3(3.5-3.0)
= .33
f= v/lambda
=343/.33m
= 1030Hz
3. what must happen in order for the sounds from the two speakers to interfere destructively, assuming that the sources are in phase?
- in order for this to happen the path difference has to be an integer number of wavelengths plus a half wavelength.
&#Good responses. Let me know if you have questions. &#