Damped Harmonic Oscillation Problem

[Chip90]

Homework Statement

A 100g mass is suspended on a rubber band with a k coefficient of 2.74 N/m. The original amplitude of the oscillations is 5cm and after 100 oscillations, the maximum speed of the weight is 0.13 m/s. Find the damping coefficient y.

Homework Equations

d2x/dt2 + γdx/dt ω02 x = 0, where γ is the damping constant

where the solution is x(t) = A exp(-γ/2t) cos(ω1 t + φ), where ω12 = ω02 -γ2/4 and A and φ depend on the initial conditions

f= 1/2pi * sqrt (m/k)

The Attempt at a Solution

I tried to use f= 1/2pi * sqrt(m/k) to find the undamped frequency to be 0.834 Hz. I then tried to use ω12 = ω02 -γ2/4 where wo is the undamped and w1 the the damped.

The problem is my teacher told me this was the wrong approach. Plus, I had no idea how to account for the 100 oscillations. I know that the Q factor is related somehow, but I am not sure how to use. Thanks!

 

[vela]

The readability of your formulas would be improved tremendously if you were to use the typographical tools the forum gives you to make subscripts and superscripts. Go into advanced mode and click on the X² and X₂ icons as needed.

Your formula for the frequency f is incorrect, but the numerical value you have is okay (though it really should be rounded to 0.833 Hz, not 0.834 Hz). You also have a typo in the differential equation.

Anyway, back to the problem. A good place to start is to see how the Q factor defined and how it is related to the damping coefficient.

 

[Chip90]

well Q= 2*pi*f/y

and Q=w₁/y=w₁/y

and w= 2*m*f

But how do 100 oscillations factor into all this?

 

[vela]

well Q= 2*pi*f/y

and Q=w₁/y=w₁/y

These two formulas are inconsistent. (I'm assuming you're using w to stand for ω and y for γ.) The first one is

Q = \frac{2\pi f}{\gamma} = \frac{\omega_0}{\gamma}

where \omega_0 = \sqrt{k/m}, but you have ω₁ in the second equation.

and w= 2*m*f

I have no idea what this equation is.

But how do 100 oscillations factor into all this?

You need to look beyond just random formulas with Q in it. What does Q represent in terms of the energy of the oscillator?

 

[Chip90]

Right I meant omega naut so that is right

that was supposed to be pi so ω= 2*\pi*f

Well Q is a dimensionless quantity, it represent the ratio of Energy stored to Energy lost. I also know that for large values of Q, Q= about number of oscillations.

Q= 2*\pi*f*\stackrel{Energy Stored}{Power Lost}

where f= undamped frequency?
Energy stored = 1/2*k*A² which is 0.003425 Joules
Power Loss ?

 

[vela]

The energy loss is, more specifically, the energy lost in one cycle. The absolute amount of energy lost per cycle decreases as the total energy of the oscillator decreases, but the fractional amount lost each cycle remains constant.

So far, you've found the initial energy, and you should be able to calculate the energy after the 100 cycles. You want to figure out at what rate the energy loss has to occur so after 100 cycles, you'll end up with the correct final energy.

 

[Chip90]

The energy loss is, more specifically, the energy lost in one cycle. The absolute amount of energy lost per cycle decreases as the total energy of the oscillator decreases, but the fractional amount lost each cycle remains constant.

So far, you've found the initial energy, and you should be able to calculate the energy after the 100 cycles. You want to figure out at what rate the energy loss has to occur so after 100 cycles, you'll end up with the correct final energy.

Ok, so is this an exponential decay? The E=E₀*e^-y*t and the energy level is 1/100 so 1/100=e^-y*t. But then I don't know the time...

or do i want the situation where the final energy is given my the velocity of the weight? In which case Efinal = 0.5 * m * v² Where E_final=0.000845 Joules?

and I also think that since Oscillations depend on f, the way to calculate for them would be Oscillations=f*t so 100=t*some frequency?

 

[vela]

The final max velocity gives you the final energy. As you found out, you generally want to avoid time because you don't know it and you can't figure it out because you don't know ω₁.

Let E be the energy of the oscillator and dE/dn be the energy lost per cycle. Then you'll have

Q = 2\pi \frac{E}{-dE/dn}

Note you have dE/dn<0 because the energy is decreasing. In the formula for Q the energy loss enters as a positive quantity, so you have to throw the minus sign in. If you integrate this equation with the appropriate limits thrown in, you can solve for Q.

 

[Chip90]

The final max velocity gives you the final energy. As you found out, you generally want to avoid time because you don't know it and you can't figure it out because you don't know ω₁.

Let E be the energy of the oscillator and dE/dn be the energy lost per cycle. Then you'll have

Q = 2\pi \frac{E}{-dE/dn}

Note you have dE/dn<0 because the energy is decreasing. In the formula for Q the energy loss enters as a positive quantity, so you have to throw the minus sign in. If you integrate this equation with the appropriate limits thrown in, you can solve for Q.

ok so I should have E= 0.003425 Joules and the integral should be from 0 to 100, which would give me a Q factor of 25.45. Then use

Q=\omega 0/\gamma to give me \gamma=0.033

 

[vela]

Hmm, you made a mistake somewhere. I found a Q factor of 449. I think you got the limits on the integrals wrong. Show me how you set it up.

 

[Chip90]

Hmm, you made a mistake somewhere. I found a Q factor of 449. I think you got the limits on the integrals wrong. Show me how you set it up.

I set it up as Q= 2* \pi \int0.003425/0.000845 dn from 0 to 100 so I think I forgot to numltiply by 100, so then I would get 2,545.. that doesn't seem right

 

[vela]

No, you can't do that. If you multiply both sides by dE/dn, you get

Q\frac{dE}{dn} = -2\pi E

Think of E as a function of n. You have a differential equation you need to solve.

 

[Chip90]

No, you can't do that. If you multiply both sides by dE/dn, you get

Q\frac{dE}{dn} = -2\pi E

Think of E as a function of n. You have a differential equation you need to solve.

ok taking the integral of that yields

Q\int\frac{dE}{dn} = -2\pi \int E

and the integral is from 0 to 100 because you want to get the energy after 100 oscillations right? but Ef is a fixed value, so how do I take an integral over this? I think it should go to this:

Q Efinal \int\frac{1}{dn} = -2\pi E

and this would give you

\frac{Q Efinal}{100} = -2pi E

 

[vela]

You should review how to solve first-order differential equations. This equation is analogous to

a \frac{dy}{dx} = -b y

where a and b are constants. This equation is separable, so how do you go about solving for y in terms of x?

 

[gneill]

There's another way to approach this without requiring that you solve the differential equation by integration. The Q is defined as:

Q = 2 \pi \frac{\text{Energy at beginning of cycle}}{\text{Energy dissipated over cycle}}

So if the initial energy is E₀, then at the end of the first cycle the energy will be:

E₁ = E₀ - E_dissipated = E₀(1 - 2π/Q)

At the end of the next cycle, it will be:

E₂ = E₁(1 - 2π/Q) = E₀(1 - 2π/Q)²

and so on...

E₃ = E₂(1 - 2π/Q) = E₀(1 - 2π/Q)³
:
:
E_n = E₀(1 - 2π/Q)ⁿ