A mass of 0.22 kg is attached to a spring and

is set into vibration with a period of 0.10 s.
What is the spring constant of the spring?
Answer in units of N/m

isn't this a standard formula? http://hyperphysics.phy-astr.gsu.edu/hbase/shm2.html

To find the spring constant (k), we can use the formula for the period of a mass-spring system:

T = 2π√(m/k)

where T is the period, m is the mass, and k is the spring constant.

Rearranging the formula to solve for k:

k = (4π²m) / T²

Given:
m = 0.22 kg
T = 0.10 s

Plugging in the values:

k = (4π² * 0.22) / (0.10)²

Calculating the expression:

k ≈ (4 * 3.14² * 0.22) / (0.10)²

k ≈ 13.9 N/m

Therefore, the spring constant of the spring is approximately 13.9 N/m.

To find the spring constant of the spring, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position. The equation is given by F = -kx, where F is the force, k is the spring constant, and x is the displacement.

In this case, the only given information is the period of vibration (T), which is the time it takes for the mass to complete one full oscillation.

The period is related to the frequency (f) by the equation T = 1/f. Thus, we can find the frequency using f = 1/T.

Given T = 0.10 s, we can calculate the frequency as follows:
f = 1 / 0.10
f = 10 Hz

Next, we know that the frequency (f) of the oscillation is related to the angular frequency (ω) by the equation ω = 2πf.

Substituting the value of frequency, we get:
ω = 2π * 10
ω = 20π rad/s

The angular frequency is also related to the spring constant (k) and the mass (m) by the equation ω = √(k/m).

Rearranging the equation to solve for k, we get:
k = ω^2 * m

Substituting the values, we get:
k = (20π)^2 * 0.22
k = 400π^2 * 0.22
k ≈ 277 N/m

Therefore, the spring constant of the spring is approximately 277 N/m.