Question

A mass of 377 g is attached to a spring and set into simple harmonic motion with a period of 0.286 s. If the total energy of the oscillating system is 6.54 ), determine the following. (a) maximum speed of the object m/s (b) force constant N/m (c) amplitude of the motion

Answer 1

a) let the maximum speed is v

maximum kinetic energy = total energy

0.50 * 0.377 * v^2 = 6.54

v = 5.89 m/s

the maximum speed of the object is 5.89 m/s

b) let the force constant is k

as maximum speed = A * w

5.89 = A * sqrt(k/0.377)

for the time period

T = 0.286 = 2pi * sqrt(0.377/k)

solving for k

k = 182 N/m

the spring constant is 182 N/m

c)

as maximum speed = A * w

5.89 = A * sqrt(k/0.377)

5.89 = A * sqrt(182/0.377)

solving for A

A = 0.268 m

the amplitude of motion is 0.268 m