A 10.6 kg object oscillates at the end of a vertical spring that has a spring constant of 2.05 × 104 N m-1
.
The effect of air resistance is represented by the damping coefficient b = 3.00 N .s m-1
. (a) Calculate the
frequency of the damped oscillation. (b) By what percentage does the amplitude of the oscillation
decrease in each cycle? (c) Find the time interval that elapses while the energy of the system drops to 5.00
% of its original value.
To answer these questions, we need to use the equation for damped harmonic motion:
m * x'' + b * x' + k * x = 0,
where:
m is the mass of the object (10.6 kg),
x is the displacement of the object from its equilibrium position,
x' is the velocity of the object,
x'' is the acceleration of the object,
b is the damping coefficient (3.00 N s/m),
and k is the spring constant (2.05 × 10^4 N/m).
(a) The equation describes an oscillation with a damping term, so the frequency of the damped oscillation can be given by the formula:
ω = √(k/m - (b/(2m))^2),
where ω (omega) is the angular frequency.
Plugging in the values, we get:
ω = √(2.05 × 10^4 N/m / 10.6 kg - (3.00 N s/m / (2 * 10.6 kg))^2).
= √(1.94 × 10^3 rad/s^2 - 0.067 N/(kg s))^2)
≈ √(1.94 × 10^3 rad/s^2 - 0.067)^2
≈ √1.879 × 10^3 rad/s^2
≈ 43.36 rad/s.
The frequency is given by:
f = ω / (2π),
≈ 43.36 rad/s / (2π)
≈ 6.90 Hz.
So, the frequency of the damped oscillation is approximately 6.90 Hz.
(b) To find the percentage by which the amplitude decreases in each cycle, we can use the formula:
γ = (b / (2m)) * 100,
where γ (gamma) is the damping factor.
Plugging in the values, we get:
γ = (3.00 N s/m) / (2 * 10.6 kg) * 100,
≈ 0.1415 * 100,
≈ 14.15%.
So, the amplitude of the oscillation decreases by approximately 14.15% in each cycle.
(c) To find the time interval that elapses while the energy of the system drops to 5.00% of its original value, we can use the formula for energy in damped harmonic motion:
E = (1/2) * k * A^2 * e^(-γt),
where E is the energy of the system, A is the initial amplitude, γ is the damping factor (as calculated in part (b)), and e is the base of the natural logarithm.
We can solve this equation for t:
ln(E / (E0 * A^2)) = -γt,
where E0 is the original energy of the system.
Plugging in the values, we get:
ln(0.05) = -0.1415t,
t ≈ (ln(0.05)) / -0.1415,
t ≈ 16.316 seconds.
So, the time interval that elapses while the energy of the system drops to 5.00% of its original value is approximately 16.316 seconds.