- #1

G01

Homework Helper

Gold Member

- 2,665

- 16

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 [tex] a = a_{max} [/tex]

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: [tex] \omega = \sqrt{\frac{k}{m}} = 5 rad/sec [/tex]

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

[tex] x=A\cos (\omega t + \phi_0) [/tex]

[tex] v_x = -\omega A\sin (\omega t + \phi_0) [/tex]

So:

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

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

a. The Amplitude

b. The Max Acceleration

c. What is the balls position when [tex] a = a_{max} [/tex]

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: [tex] \omega = \sqrt{\frac{k}{m}} = 5 rad/sec [/tex]

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

[tex] x=A\cos (\omega t + \phi_0) [/tex]

[tex] v_x = -\omega A\sin (\omega t + \phi_0) [/tex]

So:

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

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

Last edited: