The question is incomplete. The complete question is :
Two loudspeakers are placed 1.8 m apart. They play tones of equal frequency. If you stand 3.0 m in front of the speakers, and exactly between them, you hear a minimum of intensity. As you walk parallel to the plane of the speakers, staying 3.0 m away, the sound intensity increases until reaching a maximum when you are directly in front of one of the speakers. The speed of sound in the room is 340 m/s.
What is the frequency of the sound?
Solution :
Given :
The distance between the two loud speakers, 
The speaker are in phase and so the path difference is zero constructive interference occurs.
At the point
, the speakers are out of phase and so the path difference is 
Therefore,




Thus the frequency is :


Hz
Answer:
Explanation:
The time period of geosynchronous satellite must be equal to T .
The radius of its orbit will be ( R+ h )
orbital velocity V₀ = 
Time period T = 2π( R + h ) / V₀
= 2π( R + h ) x 
= R +h
h =
- R.
B is the right answer. Multiply numbers you get the answer