# Simple harmonic motion solved for time

MaxBicknell

## Homework Statement

A particle of mass 0.50 kg performs simple harmonic motion along the x axis with amplitude 0.55m and period 4.3 seconds. The initial displacement of the particle is -0.30 m and it is travelling in the positive x-direction. The phase constant of the motion (Φ) = -2.15 rad. Find the time at which the particle first reaches x = +0.3m.
Relevant equations:

## Homework Equations

x(t) = Acos(ωt + Φ)
f = 1/T = 0.2326 Hz
ω = 2πf = 1.4612 rad/s

## The Attempt at a Solution

Rearrange for t:
t = [arccos(x/A) - Φ]/ω
t = [arccos(0.6/0.55) + 2.15]/1.4612
This is giving me a maths error because arccos(x > 1) does't exist, but 0.6m is the total displacement.
Even when I use 0.3 as my measurement for x I get a ridiculous value for t:

t = [arccos(0.3/0.55) + 2.15]/1.4612
t = 40.4423 s

Not sure where I'm going wrong. Any help would be great!

Homework Helper
A particle of mass 0.50 kg performs simple harmonic motion along the x axis with amplitude 0.55m and period 4.3 seconds. The initial displacement of the particle is -0.30 m and it is travelling in the positive x-direction. The phase constant of the motion (Φ) = -2.15 rad. Find the time at which the particle first reaches x = +0.3m.
Relevant equations:

x(t) = Acos(ωt + Φ)
f = 1/T = 0.2326 Hz
ω = 2πf = 1.4612 rad/s​

My attempt:

Rearrange for t:
t = [arccos(x/A) - Φ]/ω
t = [arccos(0.6/0.55) + 2.15]/1.4612
This is giving me a maths error because arccos(x > 1) does't exist, but 0.6m is the total displacement.
Even when I use 0.3 as my measurement for x I get a ridiculous value for t:

t = [arccos(0.3/0.55) + 2.15]/1.4612
t = 40.4423 s​

Not sure where I'm going wrong. Any help would be great!
Where did the 0.6 for the displacement come from? The initial displacement was given as -0.3 m.

MaxBicknell
Where did the 0.6 for the displacement come from? The initial displacement was given as -0.3 m.

If the initial displacement is -0.3m and I am measuring the time to get to 0.3m, should x not be 0.6m?

Homework Helper
Gold Member

## Homework Statement

A particle of mass 0.50 kg performs simple harmonic motion along the x axis with amplitude 0.55m and period 4.3 seconds. The initial displacement of the particle is -0.30 m and it is travelling in the positive x-direction. The phase constant of the motion (Φ) = -2.15 rad. Find the time at which the particle first reaches x = +0.3m.
Relevant equations:

## Homework Equations

[/B]
x(t) = Acos(ωt + Φ)​
f = 1/T = 0.2326 Hz
ω = 2πf = 1.4612 rad/s

## The Attempt at a Solution

Rearrange for t:
t = [arccos(x/A) - Φ]/ω
t = [arccos(0.6/0.55) + 2.15]/1.4612
This is giving me a maths error because arccos(x > 1) does't exist, but 0.6m is the total displacement.
Even when I use 0.3 as my measurement for x I get a ridiculous value for t:

t = [arccos(0.3/0.55) + 2.15]/1.4612
t = 40.4423 s​

Not sure where I'm going wrong. Any help would be great!
A) You *have* to use x=0.3 meter. This equation gives the position as a function of time, it does not give a displacement as function of time.

B) Your calculator must be un radians to get the correct inverse cos.

MaxBicknell
A) You *have* to use x=0.3 meter. This equation gives the position as a function of time, it does not give a displacement as function of time.

B) Your calculator must be un radians to get the correct inverse cos.
Many thanks, turns out I'm just an idiot who doesn't know how to work a calculator.

Homework Helper
If the initial displacement is -0.3m and I am measuring the time to get to 0.3m, should x not be 0.6m?
No. x is displacement with respect to the origin. Better to refer it as 'position'. You know that x=Acos(wt+Φ)=0.3. w, A, and Φ are given, you can find t, but yes, set the calculator to RAD.