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:
The transverse component of acceleration is 26.32
where as radial the component of acceleration is 8.77 
Explanation:
As per the given data
u=π/4 rad
ω=u'=2 rad/s
α=u''=4 rad/s

So the transverse component of acceleration are given as

Here


So

The transverse component of acceleration is 26.32 
The radial component is given as

Here

So

The radial component of acceleration is 8.77 
Endo I think but look it up jus in case
You are correct...amplitude will be the answer