# Homework Help: Simple Harmonic Oscillator Problem

1. Jan 10, 2013

### Hibbs

1. The problem statement, all variables and given/known data

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?
2. Relevant equations
x(t)=Acos(ωt+θ)
v(t)=-Aωsin(ωt+θ)

3. 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
Please tell me where I went wrong!

2. Jan 11, 2013

### Staff: Mentor

Speed is never negative.

3. Jan 11, 2013

### rude man

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.

4. Jan 12, 2013

### ehild

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

ehild

5. Jan 12, 2013

### Staff: Mentor

I believe the formula is just misformatted and is intended to be $\cos(18s^{-1}\times t)$ - so there is no problem with the units.

6. Jan 13, 2013

### Hibbs

Thanks a lot! I got it!

Last edited: Jan 13, 2013
7. Jan 13, 2013

### rude man

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