Calculate Time to Reach Equilibrium After Spring Stretching

[bpw91284]

Homework Statement

A mass of 3.15 kg stretches a vertical spring 0.290 m. If the spring is stretched an additional 0.197 m and released, how
long does it take to reach the (new) equilibrium position again?

Homework Equations

F=-kx
T=2(pi)(m/k)^.5

The Attempt at a Solution

First I solved for k using F=kx where F is the weight of the mass

mg=kx

Then I solved for the period, T...

T=2(pi)(m/k)^.5

and the time required to reach equilibrium would be half a period so T/2 right?

 

[Doc Al]

and the time required to reach equilibrium would be have a period so T/2 right?

No. Trace out the motion of one complete period, which is the time the mass takes to return to its starting position and velocity. Hint: The initial position is y = -A; what other positions does it pass through during one period?

 

[bpw91284]

No. Trace out the motion of one complete period, which is the time the mass takes to return to its starting position and velocity. Hint: The initial position is y = -A; what other positions does it pass through during one period?

It'd go from -A, through its equilibrium position, to A, back through equilibrium, then to –A and repeat.

Did I at least solve for “k” correctly?

Also, why care about amplitude? The period has been derived and proved to be independent of amplitude.

The time required for the mass to go from state to the exact same state again (same position and velocity) is one period.

So instead of doing T/2 it’s just T right?

 

[Doc Al]

It'd go from -A, through its equilibrium position, to A, back through equilibrium, then to –A and repeat.

Exactly right. (Think about how long it takes for each of these position changes in terms of period.)

Did I at least solve for “k” correctly?

Yes. (Assuming you used the correct displacement.)

Also, why care about amplitude? The period has been derived and proved to be independent of amplitude.

This is true.

The time required for the mass to go from state to the exact same state again (same position and velocity) is one period.

Also true.

So instead of doing T/2 it’s just T right?

No. See my first parenthetical remark above.

 

[bpw91284]

Exactly right. (Think about how long it takes for each of these position changes in terms of period.)

Each position change takes T/4, and there are 4 of them...

The time required for it to go from some initial velocity and position back to the exact same velocity and position is one period.

How is that not what the problem is asking for?

 

[Doc Al]

Each position change takes T/4, and there are 4 of them...

Right.

The time required for it to go from some initial velocity and position back to the exact same velocity and position is one period.

Still true (but not what was asked).

How is that not what the problem is asking for?

Let's reread the problem:

If the spring is stretched an additional 0.197 m and released, how
long does it take to reach the (new) equilibrium position again?

You tell me: Starting position = ? End position = ?

 

[bpw91284]

You tell me: Starting position = ? End position = ?

I'm assuming when the problem states "new" equilibrium position that they mean the .290+0.197 position, which would be the starting and ending position, and it would take a period to go from start to finish.

 

[Doc Al]

I'm assuming when the problem states "new" equilibrium position that they mean the .290+0.197 position, which would be the starting and ending position, and it would take a period to go from start to finish.

The ".290+0.197 position" is the initial stretched postion, not the equilibrium position. Remember that the spring is stretched and released--so the initial position is what I called y = -A.

 

[bpw91284]

The ".290+0.197 position" is the initial stretched postion, not the equilibrium position. Remember that the spring is stretched and released--so the initial position is what I called y = -A.

Ohhhh. So it's 3/4T?

 

[bpw91284]

Also, so you think when they say "new" equilibrium position they mean the .29?

 

[Doc Al]

Ohhhh. So it's 3/4T?

Nope. Review what you wrote (about successive positions) in posts #3 and 5.

 

[Doc Al]

Also, so you think when they say "new" equilibrium position they mean the .29?

Yes, that's what they mean. (As opposed to the unstretched equilibrium position of the spring before the mass was added.)

 

[bpw91284]

Nope. Review what you wrote (about successive positions) in posts #3 and 5.

Please tell me it's 1/4T then...

 

[Doc Al]

Please tell me it's 1/4T then...

Finally...