# Thread: Solving Harmonic Motion Problems with a period/seconds in decimals

1. ## Solving Harmonic Motion Problems with a period/seconds in decimals

An object is attached to a coil spring, the object is propelled downwards fromits rest prosition at time t = 0. Write an equation for the distance of the object from its rest position after t seconds.

0 inches = distance from rest, Amplitude = 3 in, Period = 1.5 secs

I'm having problems with solving for the B in this problem because of the decimal, i was wondering how to go about doing this.

2. Originally Posted by MajorJohnson
An object is attached to a coil spring, the object is propelled downwards fromits rest prosition at time t = 0. Write an equation for the distance of the object from its rest position after t seconds.

0 inches = distance from rest, Amplitude = 3 in, Period = 1.5 secs

I'm having problems with solving for the B in this problem because of the decimal, i was wondering how to go about doing this.
$\displaystyle x = -3\cos(\omega t)$ where $\displaystyle \omega = \dfrac{2\pi}{T}$ (note that $\displaystyle T$ is the period)

3. Originally Posted by skeeter
$\displaystyle x = -3\cos(\omega t)$ where $\displaystyle \omega = \dfrac{2\pi}{T}$ (note that $\displaystyle T$ is the period)
We should really consider this to be $\displaystyle x = -3cos( \omega t + \phi)$ and state why $\displaystyle \phi = 0$. (And this isn't quite the right equation to use. Since the object is oscillating vertically we have that $\displaystyle \displaystyle x = -3cos( \omega t + \phi ) - \frac{g}{\omega ^2}$. But as this is only off by a constant we can still use simple harmonic motion by defining an x that takes this into account.)

-Dan

4. Originally Posted by topsquark
We should really consider this to be $\displaystyle x = -3cos( \omega t + \phi)$ and state why $\displaystyle \phi = 0$. (And this isn't quite the right equation to use. Since the object is oscillating vertically we have that $\displaystyle \displaystyle x = -3cos( \omega t + \phi ) - \frac{g}{\omega ^2}$. But as this is only off by a constant we can still use simple harmonic motion by defining an x that takes this into account.)

-Dan
I realize my mistake with the position of the mass at t = 0, so maybe I should have written $\displaystyle x = -3\sin(\omega t)$ since the object was "propelled downward" from its rest position at t = 0. I understood the rest position to be equilibrium ... if this is the case, then $\displaystyle -\dfrac{g}{\omega^2}$ is not really necessary.

5. Originally Posted by skeeter
I realize my mistake with the position of the mass at t = 0, so maybe I should have written $\displaystyle x = -3\sin(\omega t)$ since the object was "propelled downward" from its rest position at t = 0. I understood the rest position to be equilibrium ... if this is the case, then $\displaystyle -\dfrac{g}{\omega^2}$ is not really necessary.
You are right. I was overthinking things. The equilibrium position of the spring (which I apparently miscalculated anyway since it doesn't involve the mass of the object) would have been taken into account in the original position measurement of the spring anyway. Please ignore my previous post.

-Dan

6. chalk it up to what happens when the author of math book writes a physics problem ...

7. According to my book the answer to this problem is - 3 cos 4PI/3tThe only way i can see myself getting 4PI/3, is if i divided 2PI by 1.5 and times 2PI by 1.5 to get 3.

8. $\displaystyle \dfrac{2\pi}{1.5} \cdot \dfrac{2}{2} = \dfrac{4\pi}{3}$

9. okay thanks.