Question

An object on a horizontal, frictionless surface is attached to a spring, displaced, and then released. If it isdisplaced 0.12m from its equilibrium position and released after 0.8s its displacement is found to be 0.12m onthe opposites side, and passed the equilibrium position once during the interval. Find:

Answer 1

Given:

The maximum displacement from the equilibrium position is A = 0.12 m.

Half of the time period is

`T_{(1)/(2)}=\text{ 0.8 s}`

To find the amplitude, time period, and frequency.

Step-by-step explanation:

Amplitude is the maximum displacement from the equilibrium position.

Thus, the amplitude is A = 0.12 m.

One time period is the time taken from maximum displacement on one side(say A) to maximum displacement on the opposite side and back to the maximum displacement on the same side(A).

Thus, the time period is

`\begin{gathered} T=2T_{(1)/(2)} \\ =2*0.8 \\ =1.6\text{ s} \end{gathered}`

The frequency will be

`\begin{gathered} f=(1)/(T) \\ =(1)/(1.6) \\ =0.625\text{ Hz} \end{gathered}`