# Simple Harmonic Motion Problem

## Homework Statement

A 12.0-N object is oscillating in simple harmonic motion at the end of an ideal vertical spring. Its vertical position y as a function of time t is given by:

y(t)=4.50cmcos[(19.5s−1)t−π/8].

(a) What is the spring constant of the spring?

(b) What is the maximum acceleration of the object?

(c) What is the maximum speed that the object reaches?

(d) How long does it take the object to go from its highest point to its lowest point?

F=-kx

a=kx/m

## The Attempt at a Solution

I am unsure as to where I need to start for part a.

12N=-k(.045m) since the amplitude is the max displacement of the spring.

12N/.045m = k ==> 267N/m. However, this is wrong.

I think I'm just missing something very obvious. Any help is very appreciated!

I would begin with comparing the general equation for position with this equation

$$y(t)=4.50 cos(19.5t−\dfrac{\pi}{8})$$
$$y(t)=A cos(\omega t−\varphi)$$

We also know that $$\omega = \sqrt{\dfrac{k}{m}}$$ where k is the spring constant.

Totally forgot about the general equation that describes harmonics. Whoops.

Well, after you told me that, I figured everything out on my own. Thank you!