Simple Harmonic Oscillator Problem

Click For Summary

Homework Help Overview

The discussion revolves around a simple harmonic oscillator problem involving a mass oscillating on a Slinky. The original poster presents a formula for the position of the mass and seeks to determine its speed at a specific time.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of the velocity formula, questioning the interpretation of parameters such as angular frequency and the dimensionality of the argument in the cosine function. Some participants express confusion regarding the signs of velocity and speed.

Discussion Status

There is an ongoing exploration of the formulas and their components, with some participants providing clarifications on the interpretation of the equations. The original poster has received feedback on their calculations and the distinction between velocity and speed.

Contextual Notes

Participants note potential issues with the formatting of the equations and the assumptions made regarding the parameters involved. There is a recognition of the need for clarity in the problem statement.

Hibbs
Messages
2
Reaction score
0

Homework Statement



The position of a mass that is oscillating on a Slinky (which acts as a simple harmonic oscillator) is given by 18.5 cm cos[ 18.0 s-1t]. What is the speed of the mass when t = 0.360 s?

Homework Equations


x(t)=Acos(ωt+θ)
v(t)=-Aωsin(ωt+θ)


The Attempt at a Solution



I used the formula, v(t)=-Aωsin(ωt+θ) because you basically have everything you need such as:
A=18.5cm
ω=18.0s^-1
t=0.360s

What I get is:
v(0.360)=-(18.5cm)(18.0)sin(18.0*0.360)
to get -65.1cm/s
which isn't the right answer.
Please tell me where I went wrong!
 
Physics news on Phys.org
Speed is never negative.
 
Hibbs said:

Homework Statement



The position of a mass that is oscillating on a Slinky (which acts as a simple harmonic oscillator) is given by 18.5 cm cos[ 18.0 s-1t]. What is the speed of the mass when t = 0.360 s?

The highlighted formula makes little sense to me. First, the argument of the cos must be dimensionless but here it looks like it's time. 18.0 s must be the phase term but what's with the s? Is 18.0 in radians, deg or ? The "1" in front of t must be √(k/m), k = spring const. & m = mass, aka ω. The phase is due to the fact that this mass had initial velocity and displacement. Anyway, no way do I see that ω = 18.0.

Weird! I guess you could go

x = 18.5cos(18 - t) cm
x' = -18.5sin(18 - t) cm/s since I guess ω = 1 rad/s;
so x'(t=0.36) = -18.5sin(18 - 0.36) = -18.5sin(17.64) cm/s.

BTW v can be negative. Speed can't.
 
Hibbs said:

Homework Statement



The position of a mass that is oscillating on a Slinky (which acts as a simple harmonic oscillator) is given by 18.5 cm cos[ 18.0 s-1t]. What is the speed of the mass when t = 0.360 s?

Homework Equations


x(t)=Acos(ωt+θ)
v(t)=-Aωsin(ωt+θ)


The Attempt at a Solution



I used the formula, v(t)=-Aωsin(ωt+θ) because you basically have everything you need such as:
A=18.5cm
ω=18.0s^-1
t=0.360s

What I get is:
v(0.360)=-(18.5cm)(18.0)sin(18.0*0.360)
to get -65.1cm/s
which isn't the right answer.
Please tell me where I went wrong!

v=-65.1 cm/s is the velocity. The speed is magnitude of velocity.

ehild
 
rude man said:
The highlighted formula makes little sense to me. First, the argument of the cos must be dimensionless but here it looks like it's time.

I believe the formula is just misformatted and is intended to be [itex]\cos(18s^{-1}\times t)[/itex] - so there is no problem with the units.
 
Thanks a lot! I got it!
 
Last edited:
Borek said:
I believe the formula is just misformatted and is intended to be [itex]\cos(18s^{-1}\times t)[/itex] - so there is no problem with the units.

OK. What can be done to get the OPs to accurately state the problem, I wonder in my oft leisure moments ...
 

Similar threads

  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 18 ·
Replies
18
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
1
Views
3K
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 8 ·
Replies
8
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K