if an object on a horizontal frictionless surface is attached to a spring, displaced, and then released, it will oscillate. if it is displaced a distance 0.120 m from its equilibrium position and released with zero initial speed, then after a time 0.785 s its displacement is found to be a distance 0.120 m on the opposite side, and it has passed the equilibrium position once during this interval. Part A Find the amplitude.
Part B Find the period.
Part C Find the frequency.

The amplitude is the maximum displacement from equilibrium and it was maximum at the moment t=0.(∴The object goes from x=+A to x=-A and back to complete one cycle)

So the amplitude will be equal to the initial displacement A=0.120m

Therefore, the amplitude of the spring is 0.120m

b) movement from +A( max positive displacement) to -A (max negative displacement) took place during half the period of SHM

therefore interval by the spring will be,

0.785 s= T/2 ---> by further solving for T

T= 0.785 x 2

T=1.57 s

The period of the spring is 1.75 s

c) As we know that frequency is inversely proportional to the time period.

To find frequency of the spring:

f=1/T (∴T=1.75)

f=1/1.75

f=0.57Hz

Therefore, the frequency of the spring is 0.57 Hz with the time period of 1.75 s and amplitude 0.120 m.

To learn more about spring, refer: https://brainly.com/question/16966751

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