Simple harmonic motion solved for time

  • #1

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!
 

Answers and Replies

  • #2
ehild
Homework Helper
15,543
1,909
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.
 
  • #3
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?
 
  • #4
nrqed
Science Advisor
Homework Helper
Gold Member
3,764
291

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.
 
  • #5
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.
 
  • #6
ehild
Homework Helper
15,543
1,909
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.
 

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