Answer in units of RADIANS Two loudspeakers are placed on a wall 2.21 m apart. A listener stands directly in front of one of the speakers, 3.21 m from the wall. The speakers are being driven by a single oscillator at a frequency of 254 Hz. The speed of the sound is 343 m/s. What is the phase difference between the two waves when they reach the observer? Answer in units of rad.

phase difference = 3.21 rad  

Explanation:

given data  

wall apart =  2.21 m    

directly in front = 3.21 m

frequency f = 254 Hz  

speed of the sound v = 343 m/s

solution

we get  here first wavelength of sound wave that is  

wavelength = 343 ÷ 254  

wavelength = 1.35 m  

so  

path difference is  

path difference =  

path difference = 0.687  

path difference = 0.69 m  

and  

phase difference will be  

phase difference =    ............1  

phase difference =

phase difference = 3.21 rad

where is the photo and the answer choices?

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.