Simple Harmonic Motion of a block

[courtrigrad]

A 0.50 kg block sits on top of a 5.00-kg block that rests on the floor. The larger block is attached to a horizontal spring that has a force constant of 15.0 N/m. It is displaced and undergoes simple harmonic motion. What is the largest possible amplitude the 5.00-kg mass can have for the smaller mass to remain at rest relative to the larger block? The coefficient of static friction between the two blocks is 0.25. There is no friction between the larger block and the floor.

I have all the equations:

$$x(t) = A\cos(\omega(t) + \phi)$$
$$v(t) = -A\omega\sin(\omega(t)+\phi)$$
$$a(t) = -A\omega^{2}\cos(\omega(t)+\phi)$$
$$T = \frac{1}{f}$$
$$\omega = \sqrt{\frac{k}{m}}$$
$$T = 2\pi\sqrt{\frac{m}{k}}$$

I drew a free body diagram. I know that the velocity of the smaller block has to be 0. The force the spring exerts on the lower block has to be less than the friction force for the velocity to be 0.

Any help would be appreciated.

note: It should be $$\omega(t)$$

Thanks

 

[3trQN]

Do we not also need the dimensions of the contact surface between the two blocks? The static friction force is dependant on this information isn't it?

 

[courtrigrad]

I determined the static friction force to be $$f_{s} = (0.25)(0.5 kg)(9.8\frac{m}{s^{2}}) = 1.225 N$$

So do I just set this equal to:
$$1.225 N = (15.0 \frac{N}{m} )(x)$$, and solve for x?

thanks

 

[nrqed]

A 0.50 kg block sits on top of a 5.00-kg block that rests on the floor. The larger block is attached to a horizontal spring that has a force constant of 15.0 N/m. It is displaced and undergoes simple harmonic motion. What is the largest possible amplitude the 5.00-kg mass can have for the smaller mass to remain at rest relative to the larger block? The coefficient of static friction between the two blocks is 0.25. There is no friction between the larger block and the floor.

I have all the equations:

$$x(t) = A\cos(\omega(t) + \phi)$$
$$v(t) = -A\omega\sin(\omega(t)+\phi)$$
$$a(t) = -A\omega^{2}\cos(\omega(t)+\phi)$$
$$T = \frac{1}{f}$$
$$\omega = \sqrt{\frac{k}{m}}$$
$$T = 2\pi\sqrt{\frac{m}{k}}$$

I drew a free body diagram. I know that the velocity of the smaller block has to be 0.

No! Why would that be true??

The force the spring exerts on the lower block has to be less than the friction force for the velocity to be 0.

Any help would be appreciated.

note: It should be $$\omega(t)$$

Thanks

What you must do is to focus on the smaller block. You must have $\sum F_x = m a_x$. The only force in the x direction is the friction force (the normal force cancels gravity). Now, you know that the friction force must be smaller or equal to $\mu_s N$. Therefore, you can find the maximum acceleration...Once you know the maximum acceleration of the two blocks, you are almost there...

 

[Astronuc]

One has to determine the maximum lateral acceleration that the blocks can experience without the smaller block moving.

You found the friction force, F_s between the smaller block and the larger block. Now, the lateral for F = ma = F_s for the small block to remain stationary. With ma = $\mu\,mg$, one finds that

a = $\mu\,g$.

From the spring equation F = M a = k x, one can find that the maximum force (and therefore maximum acceleration) corresponds to the maximum displacement (amplitude).

Thus x = Ma/k, but the key here is on what mass M is the spring acting.