A mass of 0.40 kg is attached to a spring and

is set into vibration with a period of 0.15 s.
What is the spring constant of the spring?
Answer in units of N/m

what is the rearranged equation to get for k?

The period is P = 2*pi*sqrt(m/k)

You know m.

Solve for k.

[P/(2 pi)]^2 = P^2/(4*pi^2) = m/k

i go it wrong

To find the spring constant (k), we need to rearrange the equation for the period of a mass-spring system.

The formula for the period (T) of a mass-spring system is given by:

T = 2π√(m/k)

where:
T is the period in seconds
m is the mass in kilograms
k is the spring constant in N/m

By rearranging this equation, we can solve for k:

T = 2π√(m/k)
T/2π = √(m/k)
(T/2π)^2 = m/k

Now, to find the value of k, substitute the known values into the rearranged equation:

(T/2π)^2 = m/k

Given:
Period (T) = 0.15 s
Mass (m) = 0.40 kg

Substituting these values into the equation:

(0.15/2π)^2 = 0.40/k

Simplifying further:

0.0225/(π^2) = 0.40/k

Finally, to find the spring constant (k), isolate k in the equation:

k = (0.40 * π^2) / 0.0225

Evaluating the expression:

k ≈ 175.72 N/m

Therefore, the spring constant of the spring is approximately 175.72 N/m.