Find the average rate with which energy is transported along a cord by a mechanical wave of frequency f ff and amplitude A AA ,
if v v is the speed of the wave and μ μ is the mass per unit length of the cord. Express your answer in terms of the variables f ff , A AA , v vv , μ μmu , and appropriate constants.
1 answer:
You might be interested in
Answer:
ax = 6.43m/s²
Explanation:
The acceleration is the time derivative of the velocity function ax = dvx(t)/dt
We have been given the velocity function v(t) and also the velocity v = 12.0m/s and we are requested to calculate the acceleration at this time which we don't know.
So the first step is to calculate the time at which the velocity =12.0m/s and with this time calculate the acceleration. Detailed solution can be found in the attachment below.
Answer:
Explanation:
The spring is stretched by .5 m and then released that means its amplitude of oscillation A is 0.5 m .
A = 0.5 m
After the release at one extreme point , the mass comes to rest again at another extreme point after half the time period ie
T / 2 = .3 s
T = 0.6 s
Angular velocity
ω =
ω =
ω = 10.45
Maximum velocity = ω A
ω and A are angular velocity and amplitude of oscillation.
Maximum velocity = 10.45 x .5
= 5.23 m /s