1. 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!

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

Using your data, $x = 8 \cos \left(\frac{\pi}{3} t + \phi \right)$. There is insufficient information to calculate a value for $\phi$.

3. $\omega = \frac{2\pi}{T} = \frac{\pi}{3}$ rad/s

$a = -\omega^2 x$

when $x = \pm 5$ cm,

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

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

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

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

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

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

$C = 64\omega^2$

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

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

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

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

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

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

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

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

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

remember that $a$ will have a sign opposite to $x$, and $v$ can be positive or negative depending on which way the object is traveling in the cycle.

4. Thanks for that, but I'm not sure about one thing. How do I solve
at ... for t? What does the 'arccos' mean? I haven't seen that before. There are some missing pieces in my background knowledge. Cheers

5. Originally Posted by slaypullingcat
Thanks for that, but I'm not sure about one thing. How do I solve
at ... 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 $\cos^{-1}$. Read Trigonometry: Inverse Cosine and Inverse Sine - CliffsNotes