# Harmonic Oscillator

## Homework Statement

I am having issues with d) and would like to know if I did the a, b, and c correctly. I have tried to look online for explanation but with no success.

A harmonic oscillator executes motion according to the equation x=(12.4cm)cos( (34.4 rad /s)t+ π/5 ) .

a) Determine the amplitude of the oscillation.
b) Determine the maximum velocity of the oscillation.
c) Determine the period of the oscillation.
d) Determine when the object is at its equilibrium position.

max v=Aw
T=1/f

## The Attempt at a Solution

a) Determine the amplitude of the oscillation.

Amplitude would just be 12.4cm, we can take it straight out from the equation.

b) Determine the maximum velocity of the oscillation.

c) Determine the period of the oscillation.

Here we know that 2π rad is one turn so 34.4 rad is 5.48 turns.
So we have 5.48 turns per second.

T=1/f=1/5.48=0.183secs

d) Determine when the object is at its equilibrium position.

I know that the object is in equilibrium when x=0, so
0=(12.4cm)cos( (34.4 rad /s)t+ π/5 )

I may be missing some algebra skills, but I even try to compute it online and it wields no solution. What am I doing wrong?

gneill
Mentor
Your work on parts (a) through (c) looks fine.

For part (d), consider what the argument of the cosine function needs to be for the cosine to be zero. (Hint: there are many such angles)

So cos( (34.4 rad /s)t+ π/5 ) need to equal 0?

(34.4rad/s)t+π/5 has to be equal to π/2 or 3π/2?

That is going to wield t= 0.02739s and 0.1187s, does not feel right though because a period takes 0.183secs.

gneill
Mentor
So cos( (34.4 rad /s)t+ π/5 ) need to equal 0?

(34.4rad/s)t+π/5 has to be equal to π/2 or 3π/2?
Yes.

In fact, the cosine will be zero every time its argument is equivalent to π/2 or 3π/2. You should be able to write it as a function of n, where n = 0,1,2,.... Or you can solve for the first instance (n = 0 so that the argument is π/2) and then it will happen every half period after that.

alex91alex91alex