Question:

An object attached to a spring moves on a horizontal frictionless surface. This is a harmonic motion with amplitude of 0.16 m and period of 2.0 s. The mass is released from rest at \(t=0\) s and its initial displacement \(x=-0.16\) m. Find the displacement as a function of time.

Answer:

The simple harmonic motion is given as

\[  x = A\cos(\omega t + \delta)  \]

The amplitude, \(A\), is 0.16 m. Since the period \(T=2.0\) s, the frequency is calculated as

\[  f = \frac{1}{2.0}=0.50 \ \mathrm{Hz}  \]

Then, the angular frequency is

\[  \omega = 2\pi f = \pi \ \mathrm{rad/s}  \]

In order to find the phase, \(\delta\), we need to check the initial state. When \(t=0\), the wave equation becomes

\[  -0.16 = 0.16\cos(\delta)  \]

\[  \cos(\delta) = -1.0  \]

\[  \delta = \cos^{-1}(-1.0)  \]

\[  \delta = \pi  \]

Thus, we have the final form of the equation.

\[  x = 0.16\cos(\pi t + \pi)  \]