Simple Harmonic Motion Question

Click For Summary

Homework Help Overview

The discussion revolves around a particle undergoing simple harmonic motion with a specified amplitude and frequency. The original poster seeks to calculate the position, velocity, and acceleration of the particle at a given time, starting from the equilibrium position.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to apply the standard equations for simple harmonic motion but questions the correctness of their results. Some participants suggest that the choice of sine versus cosine in the equations may be a source of error, given the initial conditions.

Discussion Status

Participants have engaged in clarifying the correct form of the equations based on the initial conditions. There is an acknowledgment of the original poster's understanding of amplitude and angular frequency, and a revised approach using sine has been proposed. However, the discussion does not reach a definitive conclusion on the final calculations.

Contextual Notes

There is a focus on the implications of starting conditions for the equations used in simple harmonic motion. The original poster's reference to a textbook may indicate reliance on external sources for understanding the problem setup.

coffeem
Messages
91
Reaction score
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?
 
Physics news on Phys.org
coffeem said:

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.

coffeem said:

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.
 
rock.freak667 said:
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
 
coffeem said:
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.
 
rock.freak667 said:
Yes that should be correct now.

Thank you very much - i appreciate the hepl.
 

Similar threads

  • · Replies 51 ·
2
Replies
51
Views
4K
Replies
16
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 13 ·
Replies
13
Views
1K
Replies
5
Views
2K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
7
Views
1K