# Simple Harmonic Motion

• Apr 10th 2009, 10:44 PM
slaypullingcat
Simple Harmonic Motion
Hi, there is a question in my text that has stumped me. I have looked throught the book and there is nothing simliar in the examples. Q, The period of a particle moving in SHM has a period of 6 seconds and an amplitude of 8 cm. Calculate the velocity and acceleration when the displacement when the particle is 5 cm from the centre of motion.

How do I use p=6 s and a=8 cm to derive an equation for the displacement? Will it be cos or sin? Can it be either? How would you know?

I'm thinking I need to find the equation for the displacement making x=5 and solve for t. Then I could differentiate x to get v and a, substituting in my value for t. But how do I start?

Thanks!
• Apr 10th 2009, 11:43 PM
mr fantastic
Quote:

Originally Posted by slaypullingcat
Hi, there is a question in my text that has stumped me. I have looked throught the book and there is nothing simliar in the examples. Q, The period of a particle moving in SHM has a period of 6 seconds and an amplitude of 8 cm. Calculate the velocity and acceleration when the displacement when the particle is 5 cm from the centre of motion.

How do I use p=6 s and a=8 cm to derive an equation for the displacement? Will it be cos or sin? Can it be either? How would you know?

I'm thinking I need to find the equation for the displacement making x=5 and solve for t. Then I could differentiate x to get v and a, substituting in my value for t. But how do I start?

Thanks!

The general model is $\displaystyle x = A \cos (2 \pi f t + \phi)$: Simple harmonic motion - Wikipedia, the free encyclopedia

Using your data, $\displaystyle x = 8 \cos \left(\frac{\pi}{3} t + \phi \right)$. There is insufficient information to calculate a value for $\displaystyle \phi$.
• Apr 11th 2009, 05:28 AM
skeeter
$\displaystyle \omega = \frac{2\pi}{T} = \frac{\pi}{3}$ rad/s

$\displaystyle a = -\omega^2 x$

when $\displaystyle x = \pm 5$ cm,

$\displaystyle a = \mp \frac{5\pi^2}{9}$ $\displaystyle cm/s^2$

$\displaystyle \frac{dv}{dt} = -\omega^2 x$

$\displaystyle \frac{dv}{dt} \cdot \frac{dt}{dx} = -\omega^2 x \cdot \frac{1}{v}$

$\displaystyle v \, dv = -\omega^2 x \, dx$

$\displaystyle v^2 = -\omega^2 x^2 + C$

at $\displaystyle x = \pm 8$ cm, $\displaystyle v = 0$ ...

$\displaystyle C = 64\omega^2$

$\displaystyle v = \pm \omega \sqrt{64 - x^2}$

at $\displaystyle x = \pm$ 5 cm ...

$\displaystyle v = \pm \frac{\pi}{3} \sqrt{39}$ cm/s

or, using Mr. F's suggestion and ignoring $\displaystyle \phi$ since it doesn't matter ...

$\displaystyle x = 8\cos\left(\frac{\pi}{3} t\right)$

at $\displaystyle x = 5$ ... $\displaystyle t = \frac{3}{\pi} \arccos\left(\frac{5}{8}\right)$

use $\displaystyle t$ to evaluate $\displaystyle v$ and $\displaystyle a$ ...

$\displaystyle v = -8\left(\frac{\pi}{3}\right)\sin\left(\frac{\pi}{3} t\right)$

$\displaystyle a = -8\left(\frac{\pi}{3}\right)^2\cos\left(\frac{\pi}{ 3} t\right)$

remember that $\displaystyle a$ will have a sign opposite to $\displaystyle x$, and $\displaystyle v$ can be positive or negative depending on which way the object is traveling in the cycle.
• Apr 11th 2009, 07:17 PM
slaypullingcat
Thanks for that, but I'm not sure about one thing. How do I solve
at http://www.mathhelpforum.com/math-he...7536952b-1.gif ... http://www.mathhelpforum.com/math-he...05227fcd-1.gif for t? What does the 'arccos' mean? I haven't seen that before. There are some missing pieces in my background knowledge. Cheers
• Apr 12th 2009, 03:59 AM
mr fantastic
Quote:

Originally Posted by slaypullingcat
Thanks for that, but I'm not sure about one thing. How do I solve
at http://www.mathhelpforum.com/math-he...7536952b-1.gif ... http://www.mathhelpforum.com/math-he...05227fcd-1.gif for t? What does the 'arccos' mean? I haven't seen that before. There are some missing pieces in my background knowledge. Cheers

Use a calculator.

arcos is equivalent to inverse cos is equivalent to $\displaystyle \cos^{-1}$. Read Trigonometry: Inverse Cosine and Inverse Sine - CliffsNotes