Question

A mass is suspended on a spring. The spring is compressed so that the mass is located 5 cm above its rest position. The mass is released at time t= 0 and allowed to oscillate. It is observed that the mass reaches its lowest point 1/2s after it is released.
Required:
Find an equation that describes the motion of the mass.

Answer 1

Answer:

y = 5 cos 2πt

Explanation:

We will use the formula for simple harmonic motion curve where;

y = a cos ωt

Where;

a is amplitude

t is period

ω is angular frequency with the formula; ω = 2π/t

We are told that when the spring is compressed, the mass is located 5 cm above its rest position.

Thus;

a = 5 cm

it's highest point is 5 cm, but we are told that after 1/2 second of being released, it reaches its lowest point.

Since highest point is 5, then lowest point will be -5.

The difference in time between the highest and lowest point is ½ s. Which is half of the period.

Thus;

t/2 = ½

Thus, t = 1 s

Now, we know that;

t = 1/f = 2π/ω

Since t = 1, then 1 = 1/f

f = 1

Thus;

2π/ω = 1

Thus, ω = 2π

Thus, the equation is;

y = 5 cos 2πt