Question

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$\mathit{t}\mathbf{=}\mathbf{0}$?

Short answer

The ratio of kinetic energy to potential energy at$t=0\mathrm{is}\frac{1}{3}$

Step-by-step solution

The given data

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

2) Time,$t=0$

Understanding the concept of energy

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,$PE=\frac{1}{2}K{x}^2...............\left(\mathrm{v}\right)$

Calculation of ratio of kinetic energy to potential energy

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$\frac{1}{3}$.