Question

An air-track cart is attached to a spring and completes one oscillation every 5.67 s in simple harmonic motion. At time t = 0.00 s the cart is released at the position x = +0.250 m. What is the position of the cart when t = 29.6 s?

Answer 1

0.2447 m

Step-by-step explanation:

Amplitude, A = 0.25 m, T = 5.67 s, t = 29.6 s, y = ?

The general equation of SHM is given by

y = A Sin wt

y = 0.25 Sin (2 x 3.14 t /5.67)

Put t = 29.6 s

y = 0.25 Sin (2 x 3.14 x 29.6 / 5.67)

y = 0.2447 m