# Math Help - Harmonic Motions of Springs

1. ## Harmonic Motions of Springs

A spring of unknown spring constant is attached to a 2.30 kg mass. The mass is pulled horizontally outward by a distance "A" from equilibrium, then released at t = 0. At t = 0.200 seconds, the mass first reaches the equilibrium position, where it is seen to be moving at a velocity of v = −0.950 m/s. (a) What is the period of the motion? (b) What is the frequency of the motion? (c) Calculate the spring constant of the spring. (d) Calculate the amplitude of motion, A.

2. Originally Posted by Linnus
A spring of unknown spring constant is attached to a 2.30 kg mass. The mass is pulled horizontally outward by a distance "A" from equilibrium, then released at t = 0. At t = 0.200 seconds, the mass first reaches the equilibrium position, where it is seen to be moving at a velocity of v = −0.950 m/s. (a) What is the period of the motion? (b) What is the frequency of the motion? (c) Calculate the spring constant of the spring. (d) Calculate the amplitude of motion, A.
Try reading the first part of this.

3. I know all the equations...

4. Originally Posted by Linnus
I know all the equations...
So use them then!

Period T = 4(0.2000) = 0.8000 seconds.

f = 1/T.

$k = m (2 \pi f)^2$.

$v = -A (2 \pi f) \sin(2 \pi f t)$.

In particular, substitute v = -0.950 m/s, t = 0.2000 seconds and f = 1/T into $v = -A (2 \pi f) \sin(2 \pi f t)$ to get A. Or you could of course note that v = -0.950 m/s when t = T/4. Then $2 \pi f t = \frac{\pi}{2}$ and so $\sin (2 \pi f t) = 1$ when v = -0.950. Alternatively, you could note that when x = 0 (equilibrium position), Kinetic Energy $= \frac{k A^2}{2}$.

5. hi
how did you get the T=4(.200)? where did the 4 come from?
Thanks

6. Originally Posted by Linnus
hi
how did you get the T=4(.200)? where did the 4 come from?
Thanks
There are four stages to the motion, each stage occuring after 1/4 of a period:

From release point to equilibrium point: t = T/4.

From equilibrium to opposite position (v = 0): t = T/2.

From opposite point back to equilibrium point: t = 3T/4.