Question

A mass of 0.2kg is attached to a spring of negligible mass. If the mass executes simple harmonic motion with a period of .5 s, what will be the spring constant?

Answer 1

31.58 N/m

Explanation:

So we can find this very simply by plugging it in to the formula

`T=2pi*sqrt(m/k)`

Where,
T = Period
m = mass in kg
k = spring constant

So, plugging in numbers will give us:

`.5=2pi*sqrt(.2/k)`

If we solve for k we arrive at 31.58 Newtons per meter.