Question

A block is oscillating on a spring scale with a period of 4.60 s. At t = 0.00 s the package has zero speed and is at x = 8.30 cm. At what time after t = 0.00 s will the package first be at x = 4.15 cm?

Answer 1

Time period of oscillation, T = 4.60 s

Angular speed of te oscillating block, ω = (2 π) / T = (2π / 4.60) rad/s

Let the amplitude of oscillation is A

Then, the position of the block is

x(t) = A sin (ω t + φ) ; where φ is the phase

Now we will the given conditions to find out the position of the spring.

Condition 1: At t = 0.00 s, x = 8.30 cm

Condition 2: At t = 0.00 s, speed of the block, v = 0.00 cm/s