(Not so) Simple Harmonic Motion Problem

[shurleec]

Homework Statement

A block of mass m= 0.0750 kg is fastened to an unstrained horizontal spring whose spring constant is k= 82.0 N/m. The block is given a displacement of +0.120m, where the + sign indicates that the displacement is along the +x axis, and then released from rest.

a. What is the force (magnitude and direction) that the spring exerts on the block just before the block is released?

b. Find the angular frequency w of the resulting oscillatory motion.

c. What is the maximum speed of the block?

d. Determine the magnitude of the maximum acceleration of the block.

Homework Equations

F= -kx
x= Acos(wt)
w= 2pif
w = (k/m)^.5
v(max)= Aw
v(sho)= -Asin(wt)
a(max)= Aw^2

The Attempt at a Solution

a. I plugged the numbers into the equation, and got the answer + 9.84 N.
b. I plugged numbers into the equation w = (k/m)^.5 to get 10.5.
c. Since f = 1/t, I used w=2pif and set it equal to the answer in part b to solve for f, then got t. I use x = Acos(wt) to get amplitude. Then, I plugged everything into V(max) = Aw. The answer I got was 1.25m/s.
d. a(max)= Aw^2. I plugged numbers in and got 13.2m/s^2.

I'm sorry. I initially needed help, but figured it out as I was typing out my inquiries, so why waste all my work. I couldn't find this problem online at all so, yeah. [=

 

[tiny-tim]

Welcome to PF!

Hi shurleec! Welcome to PF!

Sometimes just typing it out for someone else to see makes it all clearer! ​

 

[nlightner]

Great job on solving the problem! Your approach and solutions seem correct. Just a couple of things to note:

- In part c, you can also use the equation v(max) = wA instead of v(max) = Aw. They are essentially the same equation, just rearranged differently.
- In part d, the units for acceleration should be m/s^2, not m/s. But the value you calculated (13.2 m/s^2) is correct.

Overall, well done on applying the equations for simple harmonic motion to solve this problem. Keep up the good work!