# Simple harmonic motion and friction

1. Apr 2, 2008

### XxBollWeevilx

1. The problem statement, all variables and given/known data

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. Relevant 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.

Last edited: Apr 2, 2008
2. Apr 2, 2008

### XxBollWeevilx

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. Apr 2, 2008

### Dick

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. Apr 2, 2008

### Dick

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. Apr 2, 2008

### XxBollWeevilx

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. Apr 2, 2008

### Dick

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

7. Apr 2, 2008

### XxBollWeevilx

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?

Last edited: Apr 2, 2008
8. Apr 2, 2008

### Dick

I think you are getting it too. But I can't quite read the conclusion. Just solve for A.

9. Apr 2, 2008

### XxBollWeevilx

Alright...thank you for the help!

10. Jun 7, 2010

### marcelot

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. Jun 8, 2010

### cartonn30gel

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.

12. Jun 8, 2010

### marcelot

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. Jun 9, 2010

### cartonn30gel

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).

Last edited: Jun 9, 2010