1. Jan 8, 2006

### G01

#1- A .3kg osscilator has a speed of .954m/s when its displacement is .03m and a speed of .714m/s when its displacement is .06m. What is the Maximum speed of the osscilator? I have no idea where to start on this one all I know are the formulas I need a hint.

Formulas: $$v_{max} = \omega A$$

$$x(t) = A\cos (\omega t + \phi_0)$$

$$v_x(t) = -\omega A\sin (\omega t + \phi_0)$$

#2- A small block (1kg) is on top of a bigger block(5kg) and they are osscilating on a spring of k= 50N/m with a period of 1.5s on a frictionless table. When the amplitude is increased to .4m the small block begins to slip. What is the coefficient of static friction in between the two blocks.

This didn't seem that hard to me. When the block just begins to slip, static friction would be eual and opposite to the force from the spring so:

$$F_{sp} = f_s$$
$$-kx = \mu_s mg$$
$$\frac{-ks}{mg} = \mu_s$$

When I fill in the numbers here i get a $$\mu_s$$ value of 2.04 which can't be right so whats wrong with my reasoning? Also I seem to be having alot of problems trying to teach myself osscilatory motion. I understand the theory well enough but doing the problems is another story altogether. If anyone has any advice or any good mathmatical methods to remember when doing these problems, please post them They'll be greatly appreciated.

2. Jan 8, 2006

### Hootenanny

Staff Emeritus
#1 - Try setting up two simulataneous equations using the velocities and displacements you are given.

3. Jan 8, 2006

### G01

Ok I set up my equations like this for #1:

$$\frac{.03}{A} = \cos\phi$$ and

$$\frac{.954}{-\omega A} = \sin\phi$$ If I square both sides and add these equations I get:

$$\frac{9X10^{-4}}{A^2} + \frac{.901}{\omega^2A^2} = 1$$

since, $$\sin^2\phi + \cos^2\phi = 1$$

Now,I do this all again with thesecond set of data and I get:

$$\frac{.0036}{A^2} + \frac{.510}{\omega^2A^2} = 1$$

Since both of these equations equal one I can set them equal to each other. When I do this I can get the $$A^2$$'s to cancel and i can solve for $$\omega$$. Then I can fill in and solve for A and then $$v_{max}$$ I still get the wrong answer. Th right answer is 1.02m/s. I get .962m/s. Is my reasoning wrong or did I just make an algebra error?

4. Jan 8, 2006

5. Jan 9, 2006

### G01

Im sorry, I just realizd that I posted wrong info for the second problem. Forget About that one I solved it all by myself!!! Sorry

6. Jan 9, 2006

### Hootenanny

Staff Emeritus
I don't think you can say they each equal each other, the sum of the two equations equal one. I think what you should get is $$\frac{.0036}{A^2} = 1 - \frac{.510}{\omega^2A^2}$$ . All your other working is right so I think you should get the right answer.

Last edited: Jan 9, 2006
7. Jan 9, 2006

### G01

Thanks Alot!

8. Jan 10, 2006

### lightgrav

In #2, I presume that the spring is attached to the bottom block,
so the top block is only accelerated by friction. The spring accelerates both blocks , so the acceleration of the top block is _______ .