Question

An object in simple harmonic motion has amplitude $8.0cm$and frequency$0.50Hz$. At $t=0s$it has its most negative position. Write the function $x\left(t\right)$that describes the object’s position.

Short answer

The function that describes the object's position is given by$x\left(t\right)=(8.0cm)\times \cos \left(\mathrm{\pi }t-\frac{\mathrm{\pi }}{2}\right)$

Step-by-step solution

Given information

amplitude $a=8.0cm$

frequency $f=0.50Hz$

Explanation

The wave equation is in the form

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

The angular frequency is in the form

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

The initial condition is that x = - a , at $t=0s$,

In order to fulfil this condition , it should satisfy the following condition:

localid="1650282632405" $\theta =-\frac{\mathrm{\pi }}{2}$

The equation now takes the form of substituting the values of angular frequency

localid="1650282642569" $x\left(t\right)=(8.0cm)\times \cos \left(\mathrm{\pi }t-\frac{\mathrm{\pi }}{2}\right)$