To answer the given questions, we need to use the equations and concepts related to damped vibrations. The differential equation that describes the motion of a damped harmonic oscillator is:
m * x'' + b * x' + k * x = 0
where:
m is the mass of the body (10 kg)
x'' is the acceleration of the body (second derivative of displacement with respect to time)
b is the damping constant (3 Ns/m)
x' is the velocity of the body (first derivative of displacement with respect to time)
k is the spring constant (2∙10^4 N/m)
x is the displacement of the body from its equilibrium position
a) To find the frequency of damped vibration, we first need to calculate the angular frequency (ω) using the formula:
ω = sqrt(k / m)
Substituting the given values:
ω = sqrt(2∙10^4 N/m / 10 kg) = sqrt(2000 rad/s)
The frequency (f) can be calculated using the formula:
f = ω / (2π)
Substituting the value of ω:
f = sqrt(2000 rad/s) / (2π) ≈ 25.23 Hz
Therefore, the frequency of damped vibration is approximately 25.23 Hz.
b) The percentage reduction in the oscillation amplitude during a period can be calculated using the formula:
Percentage reduction = (b / (2 * sqrt(k * m))) * 100
Substituting the given values:
Percentage reduction = (3 Ns/m / (2 * sqrt(2∙10^4 N/m * 10 kg))) * 100 ≈ 3.54%
Therefore, the oscillation amplitude is reduced by approximately 3.54% during a period.
c) The time taken for the energy of the system to decrease to 5% of the initial value can be determined using the equation for damped vibrations:
E(t) = E0 * exp(-2bt/m)
where:
E(t) is the energy at time t
E0 is the initial energy of the system
We want to find the time (t) when E(t) = 0.05E0.
0.05E0 = E0 * exp(-2bt/m)
Taking the natural logarithm (ln) of both sides:
ln(0.05) = -2bt/m
Solving for t:
t = (ln(0.05) * m) / (-2b)
Substituting the given values:
t = (ln(0.05) * 10 kg) / (-2 * 3 Ns/m) ≈ 0.96 seconds
Therefore, it takes approximately 0.96 seconds for the energy of the system to decrease to 5% of the initial value.