Question

A block with a mass of 0.250 kg is placed on a horizontal frictionless surface, and then attached to a spring with a spring constant of 5.00 N/m. The system is then set into motion, so that the block experiences simple harmonic motion with an amplitude of 18.0 cm.

(c) Find the smallest amount of time it takes the block to move from a position of 18.0 cm from equilibrium to a position that is just 7.00 cm from equilibrium.

Answer 1

The general equation for the displacement of the mass can be written as,
x = A cos(t)
Where A is the amplitude, A = 18 cm and is the angular frequency.
= SQRT[k/m]
Where k is the spring constant and m is the mass of the block
= SQRT[5/0.25] = 4.47 rad/s

At time t = 0,
x = A cos(0) = A = 18 cm
Consider that at time t, the displacement is 7 cm
x = A cos(t)
7 = 18 x cos(4.47 t)
cos (4.47t) = 7/18 = 0.389
4.47t = cos^-1(0.389) = 0.925            [The angle is in radians]
t = 0.925/4.47
= 0.207 s