A block of mass 2 kg hangs from a spring with constant 18 N/m. It is lifted up to a height y=20 cm and dropped from rest. a. What is the period of motion?

Question

A block of mass 2 kg hangs from a spring with constant 18 N/m. It is lifted up to a height y = 20 cm and dropped from rest.

a. What is the period of motion?

b. After 1.2 seconds where will the block be?

c. How fast will it be going at this point?

Answer 1

Let

m = 2 kg be the mass,

k = 18 N/m be the spring constant,

A = 20 cm = 0.2 m be the initial displacement,

T be the period to be calculated.

As the statement of the problem suggests, the mass moves along the y-axis,

and we consider upward direction to be positive, and downward to be negative.

Without loss of generality assume that the equilibrium position is at y = 0.

The mass will undergo harmonic motion, where at time 0 it will be at

its maximal displacement A from its equilibrium position.

By Hooke's Law at time t its position will be

y(t) = A cos(2πt/T) (1)

Its velocity, being the derivative of position, will be

v(t) = -2πA/T sin(2πt/T) (2)

a.

By Hooke's Law

T = 2π√(m/k)

Note that it is independent of A.

Substituting actual numbers:

T = 2π√(2/18) = 2π/3 s

b.

Substitute t = 1.2 into (1)

y(1.2) = 0.2 cos(2π×1.2/(2π/3)) = 0.2 cos(3.6) = -0.18 m

It will be 18 cm below the equilibrium position.

c.

Substitute t = 1.2 into (2)

v(1.2) = -2π×0.2/(2π/3) sin(2π×1.2/(2π/3)) = -0.6 sin(3.6) = 0.27 m/s

It will be moving up (positive direction) at the rate of 27 cm/s.