An object with a mass of 0.400 kg is undergoing SHM with an amplitude of 0.025 m. The maximum acceleration of the object is observed to have a magnitude of 15 m/s2. Calculate:

Question

a. the force constant of the spring

b. the maximum speed of the object

c. the acceleration (magnitude and direction) of the object when it is displaced 0.012 m to the left of its equilibrium position.

Answer 1

For all SHM

x = A sin wt

v = A w cos wt

a = - A w^2 sin wt

sin and cos has a max value of 1.

So max a = A w^2

15 = 0.025 * w^2

So w=....

w = sqrt(k/m) solve for k