67.5k views
3 votes
An underdamped oscillator of mass =1.17 kg and spring constant =32.5 N/m has a damping constant such that ^2=. What number of periods pass before the amplitude of oscillation falls to 1/^9 of the undamped amplitude? (Your answer need not be an integer.)

1 Answer

2 votes

The equation of motion for an underdamped oscillator with a damping constant b and angular frequency ω can be written as:

x(t) = e^(-bt/2m) [A cos(ωt) + B sin(ωt)]

where x(t) is the displacement of the oscillator at time t, A and B are constants determined by the initial conditions, m is the mass of the oscillator, and ω = sqrt(k/m - b^2/4m^2) is the angular frequency of the damped oscillator.

To find the number of periods that pass before the amplitude of oscillation falls to 1/(e^9) of the undamped amplitude, we can use the following equation for the amplitude of the oscillator at time t:

A(t) = e^(-bt/2m) sqrt(A^2 + B^2)

where A and B are the constants determined by the initial conditions.

Setting A(t) = 1/(e^9) A(0) and solving for t, we get:

t = (2m/b) ln(sqrt(A^2 + B^2) / (e^9) A)

We can express A and B in terms of the initial displacement x(0) and velocity v(0) as:

A = x(0)

B = (v(0) + b x(0)/(2m)) / ω

Using the given values of m, k, and b, we can calculate ω and substitute into the expression for B to get:

B = (v(0) + b x(0)/(2m)) / sqrt(k/m - b^2/4m^2)

Now we need to determine x(0) and v(0) from the initial conditions. Let's assume that the oscillator starts from rest at its maximum displacement x(0) = Amax. Then we have:

v(0) = 0

Amax = x(0) = sqrt(2E/k) = sqrt(2(0.5mv^2)/k) = sqrt(m/k)A

where E is the initial mechanical energy of the oscillator and A is the amplitude of the undamped oscillator.

Substituting these values into the expressions for A and B, we get:

A = Amax

B = bAmax/(2mω)

Substituting A and B into the expression for t, we get:

t = (2m/b) ln(sqrt(Amax^2 + (bAmax/(2mω))^2) / (e^9) Amax)

Simplifying the expression, we get:

t = (2m/b) ln(sqrt(1 + (b^2/(4m^2ω^2))) / (e^9))

Substituting ω = sqrt(k/m - b^2/4m^2), we get:

t = (2m/b) ln(sqrt(1 + (b^2/(k(m - b^2/4m^2)))) / (e^9))

Plugging in the given values of m, k, and b, we get:

t = (2(1.17)/0.45) ln(sqrt(1 + (0.45^2/(32.5(1.17 - 0.45^2/(4(1.17)))))) / (e^9)

t ≈ 6.12 periods

Therefore, it takes approximately 6.12 periods for the amplitude of oscillation to fall to 1/(e^9) of the undamped amplitude.

User Radoslav Yordanov
by
8.3k points