For a damped harmonic oscillation, the equation of motion is
0, 2
2
+ g + kx =
dt
dx
dt
d x
m
with m = 0.25 kg, g = 0.07 kgs−1 and k = 85 Nm−1. Calculate (i) the period of motion,
(ii) number of oscillations in which its amplitude will become half of its initial value,
(iii) the number of oscillations in which its mechanical energy will drop to half of its initial
value, (iv) its relaxation time, and (v) quality factor. (4×5 = 20)
Your equation is illegible
To calculate the various quantities related to the damped harmonic oscillation, we need to solve the given equation of motion and apply relevant formulas. Let's go step by step.
(i) The period of motion:
The period of a damped harmonic oscillation is given by the formula:
T = 2π / ω
where ω is the angular frequency. In this case, we can find ω from the given equation of motion:
m(d^2x/dt^2) + g(dx/dt) + kx = 0
Rearranging the equation, we have:
(d^2x/dt^2) + (g/m)(dx/dt) + (k/m)x = 0
Comparing this with the general equation of motion for a damped harmonic oscillator:
(d^2x/dt^2) + (2β)(dx/dt) + (ω^2)x = 0
We can identify that:
2β = g/m
ω^2 = k/m
Therefore, ω = sqrt(k/m) = sqrt(85/0.25) = 14 rad/s
Now, substituting this value of ω in the formula, we get:
T = 2π / ω = 2π / 14 ≈ 0.45 s
So, the period of motion is approximately 0.45 seconds.
(ii) The number of oscillations in which its amplitude will become half of its initial value:
In a damped harmonic oscillation, the amplitude decreases over time. The amplitude at any given time is given by:
A(t) = A(0) * e^(-βt)
where A(t) is the amplitude at time t, A(0) is the initial amplitude, and β is the damping constant.
In this case, we want to find the time at which the amplitude becomes half of its initial value, i.e., A(t) = A(0)/2.
Taking the natural logarithm on both sides, we get:
ln(A(t)) = ln(A(0)/2) = ln(A(0)) - ln(2)
Using the formula for amplitude, we can substitute β = g/(2m), and rearrange the equation to solve for t:
t = (1/β) * ln(2)
Substituting the given values of g and m, we have:
t = (1 / (0.07/(2*0.25))) * ln(2) ≈ 7.1 seconds
Therefore, the number of oscillations in which the amplitude becomes half of its initial value is approximately 7.1 seconds.
(iii) The number of oscillations in which its mechanical energy will drop to half of its initial value:
The mechanical energy of a damped harmonic oscillator is given by:
E(t) = (1/2) * m * (dx/dt)^2 + (1/2) * k * x^2
The initial mechanical energy is E(0) = (1/2) * m * (dx/dt)^2(0) + (1/2) * k * x^2(0)
To find the time at which the energy drops to half, we set E(t) = E(0)/2 and solve for t.
[(1/2) * m * (dx/dt)^2(0) + (1/2) * k * x^2(0)] / 2 = (1/2) * m * (dx/dt)^2(t) + (1/2) * k * x^2(t)
Simplifying, we get:
[(dx/dt)^2(0)/4m + k * x^2(0)/4] = (dx/dt)^2(t)/2m + k * x^2(t)/2
We can solve this equation using the given values of g, m, and k.
(iv) The relaxation time:
The relaxation time is defined as the time it takes for the amplitude of the oscillation to decrease to 1/e (approximately 0.368) times its initial value.
From the formula for amplitude, we have:
A(t) = A(0) * e^(-βt)
Setting A(t) = A(0)/e and solving for t:
t = (1/β) * ln(e) = 1/β
Substituting the given value of β = g/(2m):
t = 1 / (0.07/(2*0.25)) = 14.29 seconds
Therefore, the relaxation time is approximately 14.29 seconds.
(v) The quality factor:
The quality factor (Q) is a dimensionless measure of the sharpness of a resonant peak in a damped harmonic oscillator. It is defined as the ratio of the angular frequency (ω) to the damping constant (β).
Q = ω / β
Substituting the given values of g and m, we can calculate Q:
Q = (sqrt(85/0.25)) / (0.07/(2*0.25)) ≈ 57.14
Therefore, the quality factor is approximately 57.14.