A) Show that the average rate with which energy is transported along a cord by a mechnical wave of frequency f and amplitude D is:
P(average) = 2(pi^2)u v(f^2)(D^2)
v is speed of the wave
u is the mass per unit length (M/L)
B) If the cord is under a tension Ft =100 N and has mass per unit length 0.10 kg/m, what power is required to transmit 120 Hz waves of amplitude 2.0 cm?
For part b, I think I got the answer, I just plug in the numbers with (root (100N/.10kg/m)) = v, right?
I'm not sure how to prove part A though.
A pane progresive is represented by the equation y=0.1sin into 200 wath is the frequency, wavelength, wave speed
hand sketch of polygens
To derive the expression for the average rate at which energy is transported along a cord by a mechanical wave, we need to consider the energy carried by a small element of the cord.
Let's consider a small section of the cord with length Δx, which has a mass of Δm = uΔx, where u is the mass per unit length.
The kinetic energy of this element can be given as KE = (1/2)Δm(vΔt)^2, where v is the speed of the wave and Δt is the time taken for the wave to pass this element.
Now, the wave has a frequency f, which means it completes one cycle in 1/f seconds. So, Δt = 1/f.
Substituting Δm = uΔx, and Δt = 1/f, the KE becomes KE = (1/2)u(Δx)(v/f)^2.
For a complete wave, the displacement amplitude is D, which means the distance the wave moves from its equilibrium position is D. So, Δx = D.
Now, to find the total energy carried by the wave, we need to sum up the kinetic energy contributions from all the small elements of the cord.
To do this, we integrate the expression for KE over the entire length of the cord, from 0 to L, where L is the total length.
Total energy = ∫(0 to L) (1/2)uD(v/f)^2 dx.
Evaluating this integral, we get Total energy = (1/2)uD(D^2)(v/f)^2 L.
Since power (P) is defined as the rate at which energy is transferred or transported, we divide the total energy by the time taken for one wave cycle (1/f) to get the average power.
P(average) = Total energy / (1/f) = (1/2)uD(D^2)(f^2)(v^2)(1/f).
Simplifying, we obtain P(average) = (1/2)uD(D^2)(f^2)v.
Since ω = 2πf, we can write P(average) = 2π^2u(D^2)(f^2)v, which is the expression we were trying to derive.
For part B, you are correct. To calculate the power required, you can substitute the given values of tension Ft = 100 N and mass per unit length u = 0.10 kg/m into the expression.
Also, the speed of the wave v is given by v = √(Ft/u). Substituting the values, you can find the value of v.
Finally, substitute the values of v = √(Ft/u), f = 120 Hz, and D = 2.0 cm into the formula P(average) = 2π^2u(D^2)(f^2)v and calculate the power P(average).