# 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 $$\mu_s=0.60$$. What maximum amplitude of oscillation can the system have if the block is not to slip?

2. Homework Equations

$$F=-kx=ma$$

$$a=-\frac{k}{m}x$$

3. 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 $$\frac{\mu_sg}{4\pi^2f^2}$$ 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 $$\mu_smg$$ 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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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.

Dick
Homework Helper
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.

Dick
Homework Helper
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?

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.

Dick
Homework Helper
Just call the amplitude 'A'. What is the maximum acceleration of d(t)=A*sin(2*pi*f*t)?

In my book, it says that $$a=-\omega^2Acos(\omegat+\phi)$$

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

$$a=-4\pi^2f^2Acos(2ft\pi+\phi)$$

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.

$$ma=\mu_smg$$ reduces to

$$\mu_smg=a$$

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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Dick
Homework Helper
In my book, it says that $$a=-\omega^2Acos(\omegat+\phi)$$

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

$$a=-4\pi^2f^2Acos(2ft\pi+\phi)$$

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.

$$ma=\mu_smg$$ reduces to

$$\mu_smg=a$$

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.

Alright...thank you for the help!

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.

So you have
$$a=-4\pi^2f^2Acos(2\pi ft+\phi)$$

What is the maximum value that this whole thing can equal to? Think about how cosine behaves.

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)

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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