A 10.6kg object oscillates at the end of a vertical spring that has a spring constant of 2.05x104 N/m
The effect of air resistance represented by the damping coefficient . b=3N.s/m
(a) Calculate the frequency of the damping 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 initial value.
To answer these questions, we need to use the equation of motion for a damped harmonic oscillator:
m * d^2x/dt^2 + b * dx/dt + k * x = 0
where:
m = mass of the object (10.6 kg)
b = damping coefficient (3 N.s/m)
k = spring constant (2.05x10^4 N/m)
x = displacement of the object from its equilibrium position
t = time
(a) Calculate the frequency of the damped oscillation:
The frequency of the damped oscillation can be found using the formula:
f = sqrt((k/m) - (b^2/4m^2))/(2π)
Plugging in the values, we get:
f = sqrt((2.05x10^4 N/m) / (10.6 kg) - (3 N.s/m)^2 / (4*(10.6 kg)^2))/(2π)
Calculating this expression will give you the frequency of the damped oscillation.
(b) To find the percentage by which the amplitude of the oscillation decreases in each cycle, we can use the equation:
A(t) = A(0)*e^(-bt/2m)
where:
A(t) = amplitude at time t
A(0) = initial amplitude
The percentage decrease in amplitude can be calculated using the formula:
Percentage decrease = (A(0) - A(t)) / A(0) * 100
where A(t) is the amplitude at a certain time t.
(c) To find the time interval when the energy of the system drops to 5.00% of its initial value, we can use the equation for energy in a damped harmonic oscillator:
E = (1/2)kA^2
where E is the energy of the system, k is the spring constant, and A is the amplitude.
Substituting the values, we get:
E(t) = (1/2)(2.05x10^4 N/m)(A(t))^2
To find the time interval when E(t) is 5.00% of its initial value, you need to solve the equation:
E(t) = 0.05 * E(0)
(1/2)(2.05x10^4 N/m)(A(t))^2 = 0.05 * (1/2)(2.05x10^4 N/m)(A(0))^2
Solving this equation will give you the time interval.