How Do Damping Constants Affect Amplitude in Simple Harmonic Motion?

[kodee vu]

Urgent! Damped Simple Harmonic Oscillation

The thread topic says 2 masses, but there is actually only one!
I'm not asking for the complete solution for this problem; I simply just don't know WHERE to start... The question is as follows:

In Figure 15-15, a damped simple harmonic oscillator has mass m = 290 g, k = 70 N/m, and b = 75 g/s. Assume all other components have negligible mass. What is the ratio of the amplitude of the damped oscillations to the initial amplitude at the end of 20 cycles (Adamped / Ainitial)?

The figure shows a "Rigid support" at the top, to which a spring is hooked. At the bottom of this spring is a rectangular mass. At the bottom of this mass extends a vane which falls into water as the spring elongates.

variables listed: k, m, b (damping constant)
Please help if at all possible!

 

[Andrew Mason]

The thread topic says 2 masses, but there is actually only one!
I'm not asking for the complete solution for this problem; I simply just don't know WHERE to start... The question is as follows:

In Figure 15-15, a damped simple harmonic oscillator has mass m = 290 g, k = 70 N/m, and b = 75 g/s. Assume all other components have negligible mass. What is the ratio of the amplitude of the damped oscillations to the initial amplitude at the end of 20 cycles (Adamped / Ainitial)?

The figure shows a "Rigid support" at the top, to which a spring is hooked. At the bottom of this spring is a rectangular mass. At the bottom of this mass extends a vane which falls into water as the spring elongates.

variables listed: k, m, b (damping constant)
Please help if at all possible!

The solution to the differential equation of motion:

$$m\ddot x + b\dot x + kx = 0$$ is

$$x = A_0e^{-\gamma t}sin(\omega t + \phi)$$

where $\omega^2 = \omega_0^2 - \gamma^2 = k/m - b^2/4m^2$

What is the time, t, after 20 cycles? (ie. [itex]\omega t = 40\pi[/tex]?)<br /> <br /> What is $\gamma t = bt/2m$?<br /> <br /> What is the amplitude (maximum x) at this time?<br /> <br /> AM[/itex]

 

[thepatientmental]

Harmonic motion refers to the repetitive motion of an object back and forth around a central equilibrium point. In this case, we have a damped simple harmonic oscillator, which means that there is an additional force acting on the object, causing it to lose energy and gradually decrease in amplitude over time. This additional force is represented by the damping constant, b.

To solve this problem, we can use the equation for damped simple harmonic motion:

x(t) = A0e^(-bt/2m)cos(ωt + φ)

Where x(t) is the position of the object at time t, A0 is the initial amplitude, b is the damping constant, m is the mass, ω is the angular frequency, and φ is the phase angle.

First, we need to find the angular frequency, ω, using the formula ω = √(k/m), where k is the spring constant and m is the mass. Plugging in the values given in the problem, we get ω = √(70/0.29) = 13.33 rad/s.

Next, we need to find the phase angle, φ. This can be done by setting t = 0 in the equation and solving for φ. This gives us φ = 0.

Now, we can plug in the values for b, m, ω, and φ into the equation for x(t). Since we are looking for the ratio of the amplitude of the damped oscillations to the initial amplitude, we can simplify the equation to:

x(t) = A0e^(-bt/2m)

To find the amplitude at the end of 20 cycles, we need to find the position of the object at t = 20T, where T is the period of the oscillation. The period of a damped oscillator is given by:

T = 2π/ω

Plugging in the values for ω, we get T = 2π/13.33 = 0.472 s.

Therefore, at t = 20(0.472) = 9.44 s, the amplitude of the damped oscillations is given by:

Adamped = A0e^(-b(9.44)/2m)

To find the initial amplitude, we can use the fact that the amplitude of a damped oscillator decreases by a factor of e^(-πb/m) after one period. So, the