Question

A block is attached to a spring and rests on a horizontal, frictionless surface. The block is pulled to the right and released from rest when the clock starts counting and t = 0. The block oscillates back and forth about its equilibrium position, and the motion at any time is described by:x(t) = (3.5 cm)cos[(0.8 rad/s)t].What is the value of the t the second time the block goes through the equilibrium position?

Answer 1

Answer:

Explanation:

x(t) = (3.5 cm)cos[(0.8 rad/s)t]

Angular velocity ω = .8

Time period = 2π / ω

= 2 X 3.14 / .8

= 7.85 s

First time block will pass through equilibrium is T/4 ,

Second time block will pass through equilibrium

= 3.T/4

= 3/4 x 7.85

= 5.8875 s