Simple Harmonic Motion Question

  • Thread starter coffeem
  • Start date
  • #1
91
0

Homework Statement


A particle undergoes simple harmonic with amplitude of 10cm and frequency 5Hz. If it passes through the equilibrium position (x=0) at time t = 0, calculate the position, velocity and acceleration of the particle at t = 3.5m5 ms.



The Attempt at a Solution




Given that:

A = 0.1m
f = 5Hz
therefroe w = 2pif = 10pi
phase difference = 0.

So using the standard: x = Acos(wt+PD)

where PD = phase difference.

I get x = 0.1m.
differentiating this I get:
v = -0.34 m/s
and again differentiating I get:
a = -98.0 m/s^2.

However these answers appear to be wrong. Any advice on where I am messing up?
 

Answers and Replies

  • #2
rock.freak667
Homework Helper
6,223
31

The Attempt at a Solution




Given that:

A = 0.1m
f = 5Hz
therefroe w = 2pif = 10pi
phase difference = 0.

So using the standard: x = Acos(wt+PD)

where PD = phase difference.

I get x = 0.1m.
differentiating this I get:
v = -0.34 m/s
and again differentiating I get:
a = -98.0 m/s^2.

However these answers appear to be wrong. Any advice on where I am messing up?

Your attempt at getting 'A' and 'ω' are correct. But the particle starts at x=0, t=0. So your general equation should be sine and not cosine.

If it said x=A at t=0, then you'd use cosine.

Homework Statement


A particle undergoes simple harmonic with amplitude of 10cm and frequency 5Hz. If it passes through the equilibrium position (x=0) at time t = 0, calculate the position, velocity and acceleration of the particle at t = 3.5m5 ms.
 
  • #3
91
0
Your attempt at getting 'A' and 'ω' are correct. But the particle starts at x=0, t=0. So your general equation should be sine and not cosine.

If it said x=A at t=0, then you'd use cosine.


Thanks - that makes sense. I just read the position from the book without thinking...

So:

x = 0.1*sin(10pi*t)
v = 0.1*10pi*cos(10pi*t)
a = -0.1*100*pi^2*sin(10pi*t)

would that be correct? thanks
 
  • #4
rock.freak667
Homework Helper
6,223
31
Thanks - that makes sense. I just read the position from the book without thinking...

So:

x = 0.1*sin(10pi*t)
v = 0.1*10pi*cos(10pi*t)
a = -0.1*100*pi^2*sin(10pi*t)

would that be correct? thanks

Yes that should be correct now.
 
  • #5
91
0
Yes that should be correct now.

Thank you very much - i appreciate the hepl.
 

Related Threads on Simple Harmonic Motion Question

  • Last Post
Replies
3
Views
5K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
6
Views
4K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
1
Views
231
  • Last Post
Replies
10
Views
2K
Top