A damped mass-spring system oscillates at 190 Hz. The time constant of the system is 2.0 s. At t = 0, the amplitude of oscillation is 6.5 cm and the energy of the oscillating system is then 65 J.
(a) What is the amplitude of oscillation at t = 2.0 s?
What is the amplitude at t = 4.0 s?
(b) How much energy is dissipated in the first 2 s interval?
How much energy is dissipated in the second 2 s interval?
To find the amplitude of oscillation at different times and the amount of energy dissipated in different intervals, we need to understand the behavior of a damped mass-spring system.
The equation of motion for a damped mass-spring system is given by:
mx''(t) + bx'(t) + kx(t) = 0
Where:
m is the mass of the object,
x(t) is the displacement of the mass from its equilibrium position at time t,
b is the damping coefficient, and
k is the spring constant.
For a damped mass-spring system, the angular frequency of oscillation is given by:
ω = sqrt(k/m)
The natural frequency of oscillation (without damping) is given by:
ω_0 = sqrt(k/m)
The damped angular frequency is given by:
ω_d = sqrt(ω_0^2 - (b/(2m))^2)
The amplitude of oscillation decays exponentially with time, according to the equation:
A(t) = A_0 * exp(-bt/(2m))
Where:
A(t) is the amplitude at time t,
A_0 is the initial amplitude,
b is the damping coefficient, and
m is the mass of the object.
The energy of the oscillating system is given by:
E(t) = (1/2) * k * A(t)^2
Now, let's solve the given problems:
(a) What is the amplitude of oscillation at t = 2.0 s?
Using the equation for the amplitude of oscillation, we have:
A(2.0) = A_0 * exp(-bt/(2m))
Given that the time constant τ is 2.0 s, we can rewrite b as b = 2m/τ. Substituting this value, we get:
A(2.0) = A_0 * exp(-2.0t/τ)
Substituting the values given in the problem, A_0 = 6.5 cm and τ = 2.0 s:
A(2.0) = 6.5 * exp(-2.0 * 2.0 / 2.0)
Simplifying the equation, we find:
A(2.0) = 6.5 * exp(-2.0)
Calculating the value, we find:
A(2.0) ≈ 6.5 * 0.1353
Therefore, the amplitude at t = 2.0 s is approximately:
A(2.0) ≈ 0.878 cm
(b) What is the amplitude at t = 4.0 s?
To find the amplitude at t = 4.0 s, we repeat the same process as above, but substitute t = 4.0 s:
A(4.0) = A_0 * exp(-4.0 * 2.0 / 2.0)
Simplifying the equation, we find:
A(4.0) ≈ 6.5 * exp(-4.0)
Calculating the value, we find:
A(4.0) ≈ 6.5 * 0.0183
Therefore, the amplitude at t = 4.0 s is approximately:
A(4.0) ≈ 0.119 cm
(b) How much energy is dissipated in the first 2 s interval?
To calculate the energy dissipated in the first 2 s interval, we need to find the energy at t = 0 s and t = 2 s, and then subtract the two values.
Using the equation for energy:
E(t) = (1/2) * k * A(t)^2
We can calculate the energy at t = 0 s:
E(0) = (1/2) * k * A(0)^2
Given that the energy at t=0 is 65 J, we can rearrange this equation to solve for k * A(0)^2:
k * A(0)^2 = 2 * E(0) / (A(0)^2)
Substituting the values E(0) = 65 J and A(0) = 6.5 cm (converted to meters), we can calculate k * A(0)^2:
k * A(0)^2 = 2 * 65 J / (0.065 m)^2
Simplifying the equation, we find:
k * A(0)^2 ≈ 2000 N/m
Now, we can calculate the energy at t = 2 s:
E(2) = (1/2) * k * A(2)^2
Using the amplitude A(2) we calculated earlier, we can substitute this value into the equation and calculate the energy:
E(2) = (1/2) * 2000 N/m * (0.878 cm)^2
Simplifying the equation, we find:
E(2) ≈ 2000 N/m * (0.008 cm)^2
Therefore, the energy dissipated in the first 2 s interval is approximately:
E(2) ≈ 0.142 J
To find the energy dissipated in the second 2 s interval, we need to subtract the energy at t = 4 s from the energy at t = 2 s, using the same procedure as above.