What is the acceleration of a block in simple harmonic oscillation at t=3.0s?

[fstam2]

Simple Harmonic Oscillation

Hello, I am new here and wish I had found this forum earlier in the semester. Here is the situation:
A block (mass m=0.75 kg) rests on a horizontal surface (frictionless), attached to a horizontal spring (k=235 N/m). At time t=0, the block is located at the equilibrium position (x=0), given a sove that compresses the spring. The block gains a speed of 2.5 m/s instantaneously.
(ii) What is the acceleration of the block at time, t=3.0 s?

The equation I have found that might be correct is:
a= -(kA/m)sin(2(pie)t/T), where t= 3.0s and T=.352 based on
T= 2(pie)sqrt(m/k)
Using this equation I arrived at -35.2 m/s2 (the negative is because of the direction of the acceleration at t= 3.0s). This answer makes sense in that the acceleration has slowed from a max acceleration of 43.8 m/s2.

Any suggestions, advice, criticisms?
Thanks for your help.
Todd

 

[jamesrc]

Your equations are true, but I don't see how you got your answer. Like most problems, there's more than one way to do this, but considering the title of your post, consider one of the general forms for SHM of an object:

x(t) = A\sin(\omega_n t) + B\cos(\omega_n t)

where x is the position at time t, ω_n is the natural frequency, and A and B are constants determined by the initial conditions. The natural frequency in this problem is:

\omega_n = \sqrt{\frac{k}{m}}

Using the fact that x(0) = 0, we find that B = 0.

The other initial condition is the initial velocity:

\dot{x}(0) = \omega_n A \cos(\omega_n * (0)) = -2.5\,{\rm \frac{m}{s}}

(I have defined compression to be the negative x direction.)

Use this equation (with ω_n = 17.7 rad/s) to find A = -0.141 m negative just means it starts moving in compression). You can write the acceleration from the SHM equation as well (by differentiating twice):

\ddot{x}(t) = -\omega_n^2 A\sin(\omega_n t)

Solving for t = 3 s, the acceleration should be 13.2 m/s/s. You should rework the problem to see if I made a mistake, but I think I found your problem: is your calculator in degrees or radians right now? What should it be in? (I wish I noticed that before I blah-blahed the above explanation.)

 

[fstam2]

Thanks for your help. I only needed to change my calc. to radians to work out the problem. I never had the manual for the calc., and HP only has the Spanish version to download.
Todd