Question

An object in simple harmonic motion has an amplitude of $8.0cm$, a frequency of $0.25Hz$, and a phase constant of $-\mathrm{\pi }/2rad$. Draw a position graph showing two cycles of the motion.

Short answer

The position graph as follows

Step-by-step solution

Given information

amplitude $a=8.0cm$

frequency $f=0.25Hz$

phase constant $\theta =\frac{-\mathrm{\pi }}{2}rad$

number of waves$=2$

Explanation

The wave equation is in the form

$x=a\cos \left(\omega t+\theta \right)$

The angular frequency is in the form

$\omega =2\mathrm{\pi f}==2\mathrm{\pi }(0.25)=0.5\mathrm{\pi }rad/s$

The equation now takes the form of substituting the values of angular frequency and phase constant.

$x=(8.0cm)\cos \left(0.5\mathrm{\pi }t-\frac{\mathrm{\pi }}{2}\right)$

Therefore the position graph shows two cycles of motion are as follows.