Question: A damped harmonic oscillator loses 5.0 percent of its mechanical energy per cycle.
a) By what percentage does its frequency differ from the natural frequency w = sqr(k/m)?
b) After how may periods will the amplitude have decreased to 1/e of its original value?
What equations do I use? I know the final answers are 8.3 * 10^-4 percent and 39 periods. I know that 0.95A is the initial at time T. . .
Energy is directaly proportional to angular velocity,
To answer these questions, we will need to use a few equations related to a damped harmonic oscillator.
For part a) to find the percentage difference in frequency, we can use the equation:
Δω/ω = -ΔE/E,
where Δω is the difference in frequency, ω is the natural frequency, ΔE is the change in mechanical energy, and E is the initial mechanical energy.
Given that the damped harmonic oscillator loses 5.0 percent of its mechanical energy per cycle, we have ΔE/E = -0.05. Thus, we can substitute this value into the equation:
Δω/ω = -(-0.05) = 0.05.
To find the percentage difference, we multiply by 100:
Δω/ω = 0.05 * 100 = 5.0 percent.
So, the percentage difference in frequency is 5.0 percent.
For part b) to find the number of periods after which the amplitude has decreased to 1/e of its original value, we can use the equation:
A(t) = A(0) * e^(-bt/2m),
where A(t) is the amplitude at time t, A(0) is the initial amplitude, b is the damping constant, and m is the mass of the oscillator.
In this case, the initial amplitude is A(0), and after a time T, the amplitude is 0.95A(0) (given that 0.95A is the initial at time T). Therefore, we can set up the equation:
0.95A(0) = A(0) * e^(-bT/2m),
where we need to find T.
We can simplify the equation:
0.95 = e^(-bT/2m).
Take the natural logarithm of both sides to solve for T:
ln(0.95) = -bT/2m.
Rearrange the equation to solve for T:
T = -2m * ln(0.95) / b.
Now we can substitute the values we have. Given that ln(0.95) ≈ -0.051293 and that the decay constant b is related to the damping ratio ζ by b = 2ζω, we can further simplify the equation:
T = -2m * (-0.051293) / (2ζω).
The natural frequency ω = √(k/m) (given in the question). Assuming ζ is small, we can substitute ω = √(k/m) into the equation:
T = m * (-0.051293) / (ζ * √(k/m)).
Simplifying further:
T = (-0.051293 * √(m/k)) / ζ.
To find the number of periods, divide the time T by the period T0:
n = T / T0,
where T0 = 2π/ω. Substituting the values, we get:
n = (-0.051293 * √(m/k)) / (2π/ω).
Given the final answers are 8.3 * 10^-4 percent and 39 periods, you can now substitute the values back into the equations to verify the results.