Simple harmonic motion and friction

  • #1

Homework Statement



A large block P executes horizontal simple harmonic motion by sliding across a frictionless surface with a frequency f = 1.5 Hz. Block B rests on it, and the coefficient of static friction between the two is [tex]\mu_s=0.60[/tex]. What maximum amplitude of oscillation can the system have if the block is not to slip?

Homework Equations



[tex]F=-kx=ma[/tex]

[tex]a=-\frac{k}{m}x[/tex]


The Attempt at a Solution



I'm not positive that i need the equations above, but I do know that the amplitude will be given by [tex]\frac{\mu_sg}{4\pi^2f^2}[/tex] according to the textbook answer. The numerical value comes out to be 0.662 m. But I'm not sure how to associate the forces with the harmonic motion exactly. I was thinking that somehow the force must not be greater than than [tex]\mu_smg[/tex] since that is what the maximum friction force allowed would be. But I'm not sure if I need to work with the motion equations for harmonic motion or if I need to use some method involving the period, forces, etc. Any pointer in the right direction would be much appreciated.
 
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Answers and Replies

  • #2
I suppose that my two biggest things that I can't get straight in my mind are where does the pi come from, and why is there no k needed to find the answer. Also, there should be an exponent of 2 on the pi in the denominator, sorry about that.
 
  • #3
Dick
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Given the amplitude and frequency of the motion you can figure out the maximum acceleration of the block, a. Then set F=ma=mu*m*g, right? Since the frictional force must produce the needed acceleration or the block will slide.
 
  • #4
Dick
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I suppose that my two biggest things that I can't get straight in my mind are where does the pi come from, and why is there no k needed to find the answer. Also, there should be an exponent of 2 on the pi in the denominator, sorry about that.

The displacement d(t)=A*sin(2*pi*f*t) is an equation for simple harmonic motion. To get the acceleration you need to differentiate it twice. Can you see where the 4*pi^2 might come from?
 
  • #5
I see...but I don't have an amplitude, I need to find the maximum amplitude. I do see how the friction force is equal to ma though.
 
  • #6
Dick
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Just call the amplitude 'A'. What is the maximum acceleration of d(t)=A*sin(2*pi*f*t)?
 
  • #7
In my book, it says that [tex]a=-\omega^2Acos(\omegat+\phi)[/tex]

So if I put in 2*pi*f for w, I get

[tex]a=-4\pi^2f^2Acos(2ft\pi+\phi)[/tex]

OK, thanks now I can see where those quantities come from. OK, so the frictional force must not be greater than an acceleration given by this, right? I think I'm getting it now.

[tex]ma=\mu_smg[/tex] reduces to

[tex]\mu_smg=a[/tex]

I think I need to subsitute something in for a, would I put in the whole quantity above that I wrote?
 
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  • #8
Dick
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In my book, it says that [tex]a=-\omega^2Acos(\omegat+\phi)[/tex]

So if I put in 2*pi*f for w, I get

[tex]a=-4\pi^2f^2Acos(2ft\pi+\phi)[/tex]

OK, thanks now I can see where those quantities come from. OK, so the frictional force must not be greater than an acceleration given by this, right? I think I'm getting it now.

[tex]ma=\mu_smg[/tex] reduces to

[tex]\mu_smg=a[/tex]

I think I need to subsitute something in for a, would I put in the whole quantity above that I wrote?

I think you are getting it too. But I can't quite read the conclusion. Just solve for A.
 
  • #9
Alright...thank you for the help!
 
  • #10
2
0
Hello, I am stuck on this same problem and I've tried following the steps provided here, but it was not explained all the way to the end.

Here is what I've done;
a= -(omega^2)Acos(omega(t) + phi)
substituting omega = (2)(pi)(f)
a= (-4)(pi^2)(f^2)Acos[(2)(pi)(f)(t) + phi)

also, a= (umg / m) ---> a=ug

I equated these two expressions in an attempt to solve for A, but there are still several unknowns: the time t, and phase constant phi.
Any help is greatly appreciated.
 
  • #11
So you have
[tex]a=-4\pi^2f^2Acos(2\pi ft+\phi)[/tex]

What is the maximum value that this whole thing can equal to? Think about how cosine behaves.
 
  • #12
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0
That value would be 1, right? Because that is the maximum y value of cos?

Should I just equate cos[(2)(pi)(f)(t) + phi) to 1? aka, ignore it? (Sorry about my notation.. I'm new to this board so I'm not used to how to get the symbols)
 
  • #13
Yes, that's right. Cosine would oscillate between -1 and 1. So what would be the maximum value of acceleration?

Also, you don't really ignore it. It's always there; however, at some point in time, it is equal to that maximum value (or minimum).
 
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