Question

What is the phase constant for the harmonic oscillator with the velocity function$\mathbf{v}\left(t\right)$given in Figure if the position function $\mathbf{x}\left(t\right)$has the form $\mathbf{x}\mathbf{=}{\mathbf{x}}_{\mathbf{m}}\mathbf{cos}\left(\mathrm{\omega t}+\varphi \right)$? The vertical axis scale is set by ${\mathbf{v}}_{\mathbf{s}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{cm}\mathbf{/}\mathbf{s}$.

Short answer

The phase constant for the harmonic oscillator with the velocity function $\mathrm{v}\left(\mathrm{t}\right)$, if the position function has the form $\mathrm{x}\left(\mathrm{t}\right)={\mathrm{x}}_{\mathrm{m}}\cos \left(\mathrm{\omega t}+\mathrm{\varphi }\right)$, is $-0.927\mathrm{rad}$.

Step-by-step solution

Stating the given data

Vertical axis scale is set by ${\mathrm{v}}_{\mathrm{s}}=4.0\mathrm{cm}/\mathrm{s}$.

Understanding the concept of displacement equation

Using the formula of velocity, we can find the phase constant for the harmonic oscillator from theposition function of form, $\mathbf{x}\left(t\right)\mathbf{=}{\mathbf{x}}_{\mathbf{m}}\mathbf{cos}\left(\mathrm{\omega t}+\varphi \right)$.

Formulae:

The velocity of a body in motion

$\mathrm{v}=\frac{d\mathrm{x}}{d\mathrm{t}}$ (i)

Equation of displacement of the motion

$\mathrm{x}\left(\mathrm{t}\right)={\mathrm{x}}_{\mathrm{m}}\cos \left(\mathrm{\omega t}+\mathrm{\varphi }\right)$ (ii)

Calculation of phase constant of the harmonic oscillator

Differentiating equation (ii), we get

$\mathrm{v}=-{\mathrm{v}}_{\mathrm{m}}\sin \left(\mathrm{\omega t}+\mathrm{\varphi }\right)$ (iii)

Using the values$\mathrm{t}=0\mathrm{s},{\mathrm{v}}_{\mathrm{m}}=5\mathrm{cm}/\mathrm{s},\mathrm{v}=4\mathrm{cm}/\mathrm{s}$in equation (iii), we get

$\begin{array}{rcl}4\mathrm{cm}/\mathrm{s}& =& -5\mathrm{cm}/\mathrm{s}\sin \left(\mathrm{\omega }\left(0\right)+\mathrm{\varphi }\right)\\ \mathrm{\varphi }& =& {\sin }^{-1}\left(\frac{4}{5}\right)\\ & =& -0.927\mathrm{rad}\end{array}$

Therefore, the phase constant for the harmonic oscillator with the velocity function $\mathrm{v}\left(\mathrm{t}\right)$, if the position function has the form $\mathrm{x}\left(\mathrm{t}\right)={\mathrm{x}}_{\mathrm{m}}\cos \left(\mathrm{\omega t}+\mathrm{\varphi }\right)$, is $-0.927\mathrm{rad}$.