Oscillating Ball on a Spring: Analyzing Motion

[G01]

A .1 kg ball oscillates horizontally on a spring on a frictionless Table. k is 2.5 N/m. Its velocity is .2m/s when x= -.05 m What is :

a. The Amplitude
b. The Max Acceleration
c. What is the balls position when a = a_{max}

This one I know. Here x=the amplitude because a will be greatest when the spring is at its extremes.

d. What is the speed of the ball at x = .03m

So I know: \omega = \sqrt{\frac{k}{m}} = 5 rad/sec

T = 2\pi \sqrt{\frac{m}{k}} = 2/5\pi

x=A\cos (\omega t + \phi_0)

v_x = -\omega A\sin (\omega t + \phi_0)

So:

x=A \cos (\frac{2\pi t}{T} +\phi_0)

I'm lost at how to use this info to solve the problem. Any hints?

 

[Gamma]

At certain time 't' velocity and position are given. Use this on the following to get 2 equations that you can solve. x=A\cos (\omega t + \phi_0)

v_x = -\omega A\sin (\omega t + \phi_0)

Use the fact that

sin^2 \theta + cos^2 \theta = 1Accelaration can be found by differentiating v_x with respect to 't' and then proceed to find a_{max} from there.

 

[G01]

OK I've tried adding the two equations together and I get:

x+v_x=A\cos(\omega t + \phi_0) - \omega A\sin (\omega t +\phi_0)

I still don't see where to go from here. Sorry, I'm trying to teach SHM to myself and i guess i didn't do as great of a job as I thought. :)

 

[Gamma]

That is not what I said. Look at the following trig expression.

sin^2 \theta + cos^2 \theta = 1

I said use the above fact to some how get rid of sin and cosine from both of your equations

So you need to first plug in the given values for x and vx. At time t, x = -0.5 m and vx = 0.2 m/s. After you plug in these values, do the following to eliminate the trig parts from your equations. See what you get . You should be able to find the Amphlitude.

(\frac{x}{A}) ^2 + (\frac{v_x}{-wA}) ^2 = ?

 

[G01]

Ok I got the answer. Thank you very much.