Question

If an object on a horizontal frictionless surface is attached to a spring, displaced, and then released, it oscillates. Suppose it is displaced 0.125 m from its equilibrium position and released with zero initial speed. After 0.860 s, its displacement is found to be 0.125 m on the opposite side and it has passed the equilibrium position once during this interval.
Find the amplitude of the motion.

=________________m

Find the period of the motion.

=________________s

Find the frequency of the motion.

=_________________Hz

Answer 1

Answer:

a) A=0.125 m

b) T = 1.72 s

c) f= 0.58 Hz

Explanation:

a) As we are told that the maximum displacement from the equilibrium position was 0.125 m (from which it was released at zero initial speed), this is the amplitude of the resultant SHM, so, A=0.125 m

b) In order to find the period, we must get the total time needed to complete a full cycle (which means that the block must pass twice through the equilibrium point). We are told that at t=0.860 sec, the block has reached to the other end of the trajectory, and it  has passed through the equilibrium point only once.

This means that the period must be exactly the double of this time:

T = 2*0. 860 sec = 1.72 sec.

c) In a SHM, the frequency is defined just as the inverse of the period (like in a uniform circular movement), so we can get the frequency  f as follows:

f = 1/T = 1/ 1.72 s= 0.58 Hz