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
The easiest way to explain it is roughly identical to the way that your teacher explained it in class. If there were any easier way ... like writing it here in a few paragraphs ... then that's what the teacher would have done. You would have been given the easy explanation on the first day of class, printed on one sheet of paper, and you would have had the rest of the year to practice it and get really good at it.
If the class spent a month teaching it, then that's about how long it takes. Sorry.
Is recommend attaching the answer choices; Meters, Liters, Grams are three basic ones
Answer:
Explanation:
Given a particle of mass
M = 1.7 × 10^-3 kg
Given a potential as a function of x
U(x) = -17 J Cos[x/0.35 m]
U(x) = -17 Cos(x/0.35)
Angular frequency at x = 0
Let find the force at x = 0
F = dU/dx
F = -17 × -Sin(x/0.35) / 0.35
F = 48.57 Sin(x/0.35)
At x = 0
Sin(0) =0
Then,
F = 0 N
So, from hooke's law
F = -kx
Then,
0 = -kx
This shows that k = 0
Then, angular frequency can be calculated using
ω = √(k/m)
So, since k = 0 at x = 0
Then,
ω = √0/m
ω = √0
ω = 0 rad/s
So, the angular frequency is 0 rad/s
