Question

A spring hangs from the ceiling. A mass of 0.50 kg is attached to the end of the spring which then oscillates with simple harmonic motion. If the period of oscillation is measured to be 0.65 sec, what is the spring constant?

Answer 1

46.67 N/m.

Step-by-step explanation:

Using,

T = 2π[√(m/k)].............. Equation 1

Where T = period of the oscillation, m = mass attached to the end of the spring, k = spring constant of the spring.

make k the subject of the equation.

k = (4π²)m/T²................ Equation 2

Given: T = 0.65 s, m = 0.5 kg, π = 3.14

Substitute into equation 2

k = 0.5(4×3.14²)/0.65²

k = 46.67 N/m

Therefore the force constant of the spring = 46.67 N/m

Answer 2

46.77N/m

Step-by-step explanation:

to find the spring constant you can use the following formula:

`T=2\pi\sqrt{(m)/(k)}`

T: period of oscillation = 0.65s

m: mass of the object = 0.50kg

By doing k the subject of the formula and replacing you obtain:

`k=4\pi^2(m)/(T^2)=4\pi^2(0.50kg)/((0.65s)^2)=46.72(N)/(m)`

hence, the spring constant is 46.77N/m