If the phase angle for a block–spring system in SHM is $\mathit{\pi }\mathbf{/}\mathbf{6}$and the block’s position is given bylocalid="1655098514909" $\mathit{x}\mathbf{=}{\mathit{x}}_{\mathbf{m}}\mathbf{cos}\mathbf{(}\mathbf{\omega t}\mathbf{/}\mathbf{\varphi }\mathbf{)}$, what is the ratio of the kinetic energy to the potential energy at time$\mathit{t}\mathbf{=}\mathbf{0}$?

1) Phase angle for the system,$\varphi =\frac{\pi }{6}\mathrm{or}30°$

2) Time,$t=0$

Using the given angle, we can find the displacement and velocity functions. Using these functions in the equations of kinetic and potential energy, we can find the ratio between them.

Formulae:

The general expression for displacement,$x=A\cos \left(\omega t+\varphi \right)........\left(\mathrm{i}\right)$

The general expression for velocity,$v=\omega A\sin \left(\omega t+\varphi \right).............\left(\mathrm{ii}\right)$

Frequency of oscillation,$\omega =\sqrt{\frac{K}{M}}..........\left(\mathrm{iii}\right)$

Kinetic energy of a body in motion,$KE=\frac{1}{2}M{v}^2.............\left(\mathrm{iv}\right)$

Potential energy of a body in motion,$PE=\frac{1}{2}K{x}^2...............\left(\mathrm{v}\right)$

At, $t=0\mathrm{and}\mathrm{\varphi }=\frac{\mathrm{\pi }}{6}=30°$

$$\begin{gathered} x=A\cos \left(30\right)............\left(\mathrm{vi}\right) \\ V=A\omega \sin \left(30\right)..........\left(\mathrm{v}\right) \end{gathered}$$

Substituting equations (vi) & (vii) in equation (iv), we get

$$\begin{gathered} KE=\frac{1}{2}M{\left(A\omega \sin \left(30\right)\right)}^2 \\ =\frac{1}{2}M\left(A^2{\omega }^{2}0.25\right) \end{gathered}$$

Similarly substituting equations (vi) & (vii) in equation (v), we get

$$\begin{gathered} PE=\frac{1}{2}k{\left(A\cos \left(30\right)\right)}^2 \\ =\frac{1}{2}k\left(A^{2}0.75\right) \\ =\frac{1}{2}{\omega }^{2}M\left(A^{2}0.75\right) \end{gathered}$$

The ratio of KE to PE will be,

$$\begin{gathered} \frac{KE}{PE}=\frac{{\displaystyle \frac{1}{2}}M\left(A^2{\omega }^{2}0.25\right)}{{\displaystyle \frac{1}{2}}{\omega }^{2}M\left(A^{2}0.75\right)} \\ =\frac{1}{3} \end{gathered}$$

Hence, the value of the ratio is$\frac{1}{3}$.