Question

An object in simple harmonic motion has amplitude $4.0cm$and frequency $4.0Hz$, and at $t=0s$it passes through the equilibrium point moving to the right. Write the function$x\left(t\right)$ that describes the object’s position

Short answer

The object's position function$x\left(t\right)$at $x\left(t\right)=(4.0cm)(\cos \left(8\mathrm{\pi }t-\mathrm{\pi }/2\right)$.

Step-by-step solution

Step: 1 Given data: 

amplitude $a=4.0cm$

frequency $f=4.0Hz$

The pulse equation is written in the following way:

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

The having an appropriate is calculated in the following way:

$\omega =2\mathrm{\pi f}=2\mathrm{\pi }\left(4\right)=8\mathrm{\pi }rad/s$

Step: 2 Explanation:

The item travels through all the equilibria to the right, with the initial state at $t=0s$.

$x(t=0s)=a\cos \left(\theta \right)=0$

The cosine term must satisfy the following criterion in order to be zero:

$\theta =\frac{n\mathrm{\pi }}{2}$

Because the item is travelling in a positive direction, its velocity at $t=0s$ is positive. The gradient of the orientation function is the velocity function:

$$\begin{gathered} v\left(t\right)=-\omega a\sin (\omega t+\theta ) \\ v(t=0s)=-\omega a\sin \left(\theta \right)>0 \end{gathered}$$

Then$\sin \theta$ must be negative, and $\theta$ must also be negative.

Converting the value of angular frequency into the equation now completes the equation.

$x\left(t\right)=(4.0cm)\times \cos \left(8\mathrm{\pi }t-\frac{\mathrm{\pi }}{2}\right)$