Calculating Amplitude for a Spring Oscillation with Hooke's Law

[Dark Visitor]

I attach a 2 kg block to a spring that obeys Hooke's Law and supply 16 J of energy to stretch the spring. I release the block; it oscillates with period .3 s. The amplitude is:


* 38 cm
* 19 cm
* 9.5 cm
* 4.3 cm


Can anybody help me with this? I can't figure out what to do or where to start. It seems so easy, but I can't get it. Thanks.

 

[denverdoc]

Do you recall seeing an eqn T=2*pi*sqrt[k/m] . I think that is it. So solve for K by squaring both sides. Recall that energy is 1/2 k x^2 and is equal to 16J. Solve for x. This will be the initial stretch. In SHM, this constant represents 1/2 amplitude.

 

[Delphi51]

Could you list the spring formulas you have to work with? There should be one with force, one with energy and one with period or frequency.

 

[Dark Visitor]

To Denverdoc: Yes, but on my notes, I have:

T = 2$$\pi$$*$$\sqrt{}m/k$$



To Delphi51: I will list the ones I see here with me.

Potential Energy: U_sp = 1/2(k)(x_s)²
(k is spring constant, x_s is compression/extension of spring from relaxed position)

F_s = -kx_s

W_{by sp} = 1/2(k)(x_s,i)² - 1/2(k)(x_s,f)²

Those are the only ones I see at the moment.

 

[denverdoc]

Yes you're right I forgot to flip it in going from w to T. So you have enough to solve it. Just follow the instructions in my last post.

 

[Dark Visitor]

It's all good. I do have a question: You said in your last post that to solve for K, I had to square both sides. It seems to me that I have to do much more than that. Am I right?

 

[denverdoc]

Sure, there is some simple algebra as well. Remember that is to just get K. You still have the other things to do as outlined in my post.

 

[Dark Visitor]

Well, when solving for k, I got:

k = m/(t/2$$\pi$$)²

is that right?

 

[denverdoc]

Yes (but by convention capital T is used for the period to avoid confusion with time.)

 

[Schrodinger's Dog]

To Denverdoc: Yes, but on my notes, I have:

T = 2$$\pi$$*$$\sqrt{}m/k$$



To Delphi51: I will list the ones I see here with me.

Potential Energy: U_sp = 1/2(k)(x_s)²
(k is spring constant, x_s is compression/extension of spring from relaxed position)

F_s = -kx_s

W_{by sp} = 1/2(k)(x_s,i)² - 1/2(k)(x_s,f)²

Those are the only ones I see at the moment.

He's right make sure T=t in your terms or you could end up with anything.

 

[Schrodinger's Dog]

Yes (but by convention capital T is used for the period to avoid confusion with time.)

Exactly right.

 

[Dark Visitor]

Yeah, I just missed hitting the shift key when typing, or I typed it too fast. I do know the difference, and wouldn't get confused.

Well, I did that and got

k = 2 kg/(.3 s/2$$\pi$$)²
= 9.006

 

[denverdoc]

Very well. I like that it is round. Now it's perfectly ok to go back and reread my earlier posts as to how to proceed, remember we are looking for x, the initial displacement which is 1/2A.

 

[Schrodinger's Dog]

Very well. I like that it is round. Now it's perfectly ok to go back and reread my earlier posts as to how to proceed, remember we are looking for x, the initial displacement which is 1/2A.

Again he's right more mistakes come from mixing up terms than everything else.

[tex]a\neq{x}[/itex]

[tex]c\neq{C}[/itex]

And so on.

Yeah, I just missed hitting the shift key when typing, or I typed it too fast. I do know the difference, and wouldn't get confused.

Well, I did that and got

k = 2 kg/(.3 s/2$$\pi$$)²
= 9.006

Looks good to me?

What expectation value were you expecting for k?

 

[Dark Visitor]

Okay. Now I solved for x:

E = 1/2(k)(x)²
16 J = 1/2(9.006)(x)²

x = $$\sqrt{}(32 J/9.006)$$
= 1.885

 

[Dark Visitor]

So that means that 2x is my amplitude? so all I need to do is double my x and get my final amplitude value?

 

[Schrodinger's Dog]

So that means that 2x is my amplitude? so all I need to do is double my x and get my final amplitude value?

Indeed. Sounds like you have this all figured out gj.

 

[denverdoc]

If the question is looking for peak to peak amplitude, yes. Remember your cos or sin function will vary between 1 and -1.

The initial displacement becomes the A on the outside of the SHM type of eqns you have been writing this weekend. Since it is simple harmonic motion, the initial displacement determines the subsequent behavior. Had we stretched the string more initially, the A would be different. Imagine a pendulum. If there is no friction, the ball swings back and forth to and from ther release point--same principle here.

 

[Schrodinger's Dog]

If the question is looking for peak to peak amplitude, yes. Remember your cos or sin function will vary between 1 and -1.

The initial displacement becomes the A on the outside of the SHM type of eqns you have been writing this weekend. Since it is simple harmonic motion, the initial displacement determines the subsequent behavior. Had we stretched the string more initially, the A would be different. Imagine a pendulum. If there is no friction, the ball swings back and forth to and from ther release point--same principle here.

For simple harmonic operators this might help? Don't know?

https://www.physicsforums.com/showthread.php?t=219445&highlight=Wilson-Sommerfield

Helped me though.

 

[Dark Visitor]

I think I understand. My only issue now is, when I got 3.77 m as my A, how is the answer 38 cm? Did I mess up somewhere with my math?

 

[Schrodinger's Dog]

I think I understand. My only issue now is, when I got 3.77 m as my A, how is the answer 38 cm? Did I mess up somewhere with my math?

Order of magnitude?

http://en.wikipedia.org/wiki/Order_of_magnitude

ttp://upload.wikimedia.org/math/2/e/1/2e144dcb120e3ba404112929f2ff82f5.png

what's the answer to x dp?

 

[Dark Visitor]

I don't know. I got 3.77 m, which converts to 377, or 380, cm. But my answer is 38 cm. I don't see what I did wrong...

 

[Dark Visitor]

So what do I do? I can't find where my error is...

 

[denverdoc]

i found it--it is in the algebra for computing K. You have it flipped a bit, 2 pi should be in the numerator. k=219.32 when you do this, you will get .38m but it isn't doubled. It is not asking for peak to peak amplitude but just one leg. Sorry, I should have checked that more carefully and the round number lulled me into a false sense of security.

 

[Redbelly98]

Mod note:
Deliberate attempts to confuse the OP, and responses, have been deleted.

 

[Dark Visitor]

Why is 2$$\pi$$ in the numerator? That's not the equation I had written down...

I do have another one that says:

T_spring/mass = 2$$\pi$$/$$\sqrt{}k/m$$

 

[denverdoc]

thanks.

 

[Dark Visitor]

Why thanks? And why would we use that one now?

 

[denverdoc]

T=2*pi*(sqrt (m/k)) Remember the rule about dividing being equal to mult by the reciprocal:

a/b/c/d= ad/bc.

 

[denverdoc]

Why thanks? And why would we use that one now?

Sorry--the thanks was to the moderator for removing some confusion. But this whole thread seems hopelessly confused. Nevertheless, do the algebra carefully and you will get an answer for x=0.38m which is what we wanted!

 

[Dark Visitor]

Okay, I redid it using the new equation. I think I got the same things you did, which is good. :) But my question is: are we doubling it for the same reason we did before? (A = 2x)

 

[denverdoc]

Okay, I redid it using the new equation. I think I got the same things you did, which is good. :) But my question is: are we doubling it for the same reason we did before? (A = 2x)

There are many kinds of amplitude: peak-to-peak and what is being used here as in

A cos(wt). Apparently they just want A. Whew glad that's over. I'm going to leave you in the good hands of whomever for the last problem before I kill us both with confusion.

BTW, the equation is the same, just more legible by taking out the extra divisor sign which is what led to the bunged algebra. Don't feel bad, do it all the time myself.

 

[Dark Visitor]

Okay, I think I can get that one eventually. There is a guy on there helping me now, so thanks to both of you for your help. I know it was a pain in the butt. But I appreciate it.