How do I model this function? (Damped Harmonic Motion)

[daisy7777]

Homework Statement

A spring with an upstretched length of 50.0 cm and a spring constant of 120.0 N/m is suspending a 1.50 kg mass below the ceiling. if the mass is pulled down 10.0 cm from its equilibrium position and released we notice that the vibration has a half life of 1.40 s. What is the position of the mass at t = 2.00 s?

Relevant Equations

Fe = Fg
x = Δx(1/2)^t/λ*cos(sqrt(k/m)*t)+c

I first solved for the extension of the spring when its at equilibrium w/ the mass. I got Δx = 0.122m. I thought that if I added this with 0.1m, this would give me my amplitude. I then set 0.5m as my c value and plugged the rest of my values in from there. What am I doing wrong?

 

[haruspex]

I thought that if I added this with 0.1m, this would give me my amplitude

No. Think about the motion. When the mass passes its initial position on its way up, which way is the net force on it?

 

[daisy7777]

No. Think about the motion. When the mass passes its initial position on its way up, which way is the net force on it?

It's negative, right?

 

[haruspex]

It's negative, right?

Right. Do you see a symmetry there? Compare the net force when x above the initial (equilibrium) position with that when x below it.

 

[kuruman]

Also note that if you write the underdamped equation as $$\ddot x+2\beta \dot x +\omega_0^2~x=0,$$ the frequency in the phase of the sinusoidal should not be the natural frequency ##\omega_0=\sqrt{k/m}~## but ##\omega=\sqrt{\omega_0^2-\beta^2}.##

 

[daisy7777]

Also note that if you write the underdamped equation as $$\ddot x+2\beta \dot x +\omega_0^2~x=0,$$ the frequency in the phase of the sinusoidal should not be the natural frequency ##\omega_0=\sqrt{k/m}~## but ##\omega=\sqrt{\omega_0^2-\beta^2}.##

That's the equation my teacher's given us, so I'm assuming it's the one he wants us to use for problems like this. Thank you though!

 

[daisy7777]

Right. Do you see a symmetry there? Compare the net force when x above the initial (equilibrium) position with that when x below it.

So my net force would be equal to Fs = Δx*k in both directions, right? I'm writing it out now and I get a magnitude of 60N in both directions as I'm using 50cm as Δx since that's the length when the spring is upstretched, would this be correct? And if so, would I have to consider the value I got for Δx in Δx*k = mg as the c value instead?

 

[nasu]

What is the equation for the displacement versus time given in your course? Didn't have an exponentially decaying amplitude?

 

[haruspex]

So my net force would be equal to Fs = Δx*k in both directions, right? I'm writing it out now and I get a magnitude of 60N in both directions as I'm using 50cm as Δx since that's the length when the spring is upstretched, would this be correct? And if so, would I have to consider the value I got for Δx in Δx*k = mg as the c value instead?

I think you mean unstretched, not upstretched.
When the mass is attached to the unstretched spring and lowered gently to its equilibrium position, the spring stretches by ##x_1## where ##kx_1=mg##.
The mass is then pulled down an additional ##\Delta x=0.1##m for a total stretch of ##x_1+\Delta x##.
If there were no losses, when released it would execute SHM symmetrically about the equilibrium position, not the unstretched position. So what is the initial amplitude?

 

[daisy7777]

I think you mean unstretched, not upstretched.
When the mass is attached to the unstretched spring and lowered gently to its equilibrium position, the spring stretches by ##x_1## where ##kx_1=mg##.
The mass is then pulled down an additional ##\Delta x=0.1##m for a total stretch of ##x_1+\Delta x##.
If there were no losses, when released it would execute SHM symmetrically about the equilibrium position, not the unstretched position. So what is the initial amplitude?

Would it be 0.1m + 0.122m?

 

[kuruman]

Look at the expression your teacher gave you.
Clearly, ##\Delta x## is the displacement of the mass from the equilibrium position.
Variable ##x## on the left-hand side is the position of the mass. Position requires an origin. Where is the origin here, at the unstretched end of the spring or the equilibrium position?

 

[daisy7777]

Look at the expression your teacher gave you.
Clearly, ##\Delta x## is the displacement of the mass from the equilibrium position.
Variable ##x## on the left-hand side is the position of the mass. Position requires an origin. Where is the origin here, at the unstretched end of the spring or the equilibrium position?

Wouldn't it be at the unstretched end?

 

[haruspex]

No.
At the equilibrium position, ##k x_{eq}=mg##. The net force on the mass is zero.
At any downward displacement from there, to ##x_{eq}+\Delta x##, the net upward force is##k(x_{eq}+\Delta x)-mg=k\Delta x##.
This means you can forget about mg and ##x_{eq}## because they always cancel. What you are left with is a displacement of ##\Delta x## from the equilibrium position and no gravity.

 

[daisy7777]

No.
At the equilibrium position, ##k x_{eq}=mg##. The net force on the mass is zero.
At any downward displacement from there, to ##x_{eq}+\Delta x##, the net upward force is##k(x_{eq}+\Delta x)-mg=k\Delta x##.
This means you can forget about mg and ##x_{eq}## because they always cancel. What you are left with is a displacement of ##\Delta x## from the equilibrium position and no gravity.

Ok, so then would my Δx just be 10cm as that's the displacement from equilibrium that the spring is pulled and therefore it would be my amplitude?

 

[haruspex]

Ok, so then would my Δx just be 10cm as that's the displacement from equilibrium that the spring is pulled and therefore it would be my amplitude?

Yes, that is the initial amplitude.

 

[daisy7777]

Yes, that is the initial amplitude.

So, would my c value be the Δx between the string and mass then? The 0.122m I calculated earlier? I'm so sorry for the myriad of questions, it's just I haven't had much practice w/ modelling damped SHM functions.

 

[kuruman]

What is the equation for the displacement versus time given in your course? Didn't have an exponentially decaying amplitude?

The given expression does have an exponentially decaying amplitude of sorts. Note that $$\left(\frac{1}{2}\right)^{t/\lambda}=2^{-t/\lambda}.$$ The teacher is using 2 for the base instead of 2.7182818##\dots~~## which would be OK for ##x(t)## but finding ##\dot x(t)## might require some extra attention.

 

[kuruman]

So, would my c value be the Δx between the string and mass then?

That would depend on where you choose to put the origin for the position of the mass as indicated in post #11. Whatever point you choose as this origin, when you evaluate your teacher's expression at ##t=0##, the mass has to be a distance of 10.0 cm below the equilibrium position.

 

[daisy7777]

That would depend on where you choose to put the origin for the position of the mass as indicated in post #11. Whatever point you choose as this origin, when you evaluate your teacher's expression at ##t=0##, the mass has to be a distance of 10.0 cm below the equilibrium position.

Ok so I got this to be my equation, however, it yields an answer that doesn't match the one in the answer key which is 65.8cm. I'm getting 0.0213m. Is there something I'm doing wrong? At t = 0s, the mass is 10cm above the origin.

 

[kuruman]

At t = 0s, the mass is 10cm above the origin.

That places the origin an additional 10 cm below the equilibrium position.

Ok so I got this to be my equation, however, it yields an answer that doesn't match the one in the answer key which is 65.8cm.

Try placing the origin with respect to which you measure the position of the mass at the fixed end of the spring. That's the only way to get something as large as 65.8 cm.