Question

A mass of 0.38 kg is attached to a spring and is set into vibration with a period of 0.19 s. What is the spring constant of the spring? Answer in units of N/m

Answer 1

Spring constant, K = 415.9 N/m

Step-by-step explanation:

The period of vibration of a spring, T, is given by:

T = 2√m/K

Where m = mass of the body in kg, K = spring constant or force constant or elastic constant or stiffness in Nm-1

Making K the subject of the expression,

K = m(2)2/(T)2

K = 4m()2/(T)2

Where m = 0.38kg, = 3.14285714286, T = 0.19s

K = 4 x 0.38 (3.14285714286)2 / (0.19)2

K = 4 x 0.38 x 9.87755102043 / 0.0361

K = 15.0138775511 / 0.0361

K = 415.9 Nm-1