# Oscillations Quick Question

A graph resembling that of cos(x) is presented in which x is the vert. axis and t is the horizontal axis. Assume that the x coordinate of point R is 0.12 , and the t coordinate of point K is 0.0050.

So . . .
R = (0 , 0.12 m)
K = (.005 s ,0)

What is the period T of oscillations?

I know that a*T = .005 s, where T is the period and a is the fraction of a full wavelength covered over this interval. What is a so I can get T?

Thanks for any hints.

Have you tried using kinematics? You might be able to use the formula d = vit+ ½ at² to find "a" if you mean "a" as in 'acceleration'.

Gokul43201
Staff Emeritus
Gold Member
ms. confused said:
Have you tried using kinematics? You might be able to use the formula d = vit+ ½ at² to find "a" if you mean "a" as in 'acceleration'.
That wouldn't be the right way to do it.

Soaring Crane, you haven't sufficiently described the figure.

Is R at the topmost point (like it would be for cosx) ?
And is K the point where the curve first intersects the t-axis ?
Or are they somethings else ?

Code:
R?
*   *
|        *
|           *
|____________ *K?____
|              *