Two loudspeakers, 4 meters apart and facing each other, play identical sounds of the same frequency. You stand halfway between them, where there is a maximum of sound intensity. Moving from this point towards one of the speakers, you encounter a minimum of sound intensity when you have moved 0.25 meters. Velocity of the sound is 343 m/s
A. What is the frequency of the sound?
B. If the frequency is then increased while you remain 0.35m from the center, what is the first frequency for which that location will be a maximum of sound intensity?
m/s

Explanation:

Minimum intensity occurs due to destructive interference of sound. For it to take place ,

path difference = odd multiple of half wave length

When moved by .025 m ,

path difference created = ( x + .25 ) - ( x - .25 )

= 2 x .25

= .5 m

So .5 = half wave length

wave length = 2 x .5

= 1 m

frequency = velocity / wave length

= 343 / 1

= 343 Hz

B )

Now when frequency is increased , wave length will be decreased . For maximum intensity , constructive interference will have to take place .

For that

path difference = integral multiple of wave length

0.5 = 1 x wave length

wave length = .5

frequency = 343 / .5

= 686 Hz

momentum of soft ball three times less than baskey ball