Question

What is the phase constant for the harmonic oscillator with the position function$\mathit{x}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$given in Figure if the position function has the form$\mathbf{x}\mathbf{=}{\mathbf{x}}_{\mathbf{m}}\mathbf{cos}\mathbf{(}\mathbf{\omega t}\mathbf{+}\mathbf{f}\mathbf{)}$? The vertical axis scale is set by${\mathbf{x}}_{\mathbf{m}}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{cm}$.

Short answer

Phase constant for motion is 1.91 rad

Step-by-step solution

The given data

Vertical axis scale is${\mathrm{x}}_{\mathrm{m}}=6\mathrm{cm}$ .

Understanding the concept of phase

Phase is the cosine of an angle. We can use this concept to find the phase. We are given the position versus time graph from which we find the position at t=0 sec. The amplitude is given. So, we use that to find the phase.

Calculation of phase constant

We are given the graph of position vs time. In that at $t=0$position is x=--2cm and amplitude is 6 cm so, the phase is calculated as follows:

We know the equation of position,

$x=x_m\cos \left(\omega t+f\right)$

$$\begin{gathered} -2cm=\left(6cm\right).\cos \left(\omega t+f\right) \\ f+\omega t=1.91\mathrm{rad} \end{gathered}$$

At,t=0, we have the phase constant of the oscillation as,$\varphi =1.91\mathrm{rad}$.

Therefore, the phase constant is 1.91 rad .