Question

A particle undergoing simple harmonic motion has time-dependent displacement given by $x\left(t\right)=Asin\frac{\pi t}{90}$. The ratio of kinetic to the potential energy of this particle at $t=210s$ will be

• A. $1$
• B. $\frac{1}{3}$
• C. $\frac{1}{9}$
• D. $2$

Answer 1

The correct option is B

$\frac{1}{3}$

Step 1: Given Data:

$\Rightarrow \omega =\frac{\mathrm{\pi }}{90}$

$\Rightarrow t=210s$

Step 2. The formulae of the energies of a particle in simple harmonic motion are given as:

Kinetic Energy, $\begin{array}{l}k=\frac{1}{2}m{\omega }^2{A}^{2}co{s}^2\omega t\end{array}$

Potential Energy,$\begin{array}{l}U=\frac{1}{2}m{\omega }^2{A}^{2}si{n}^2\omega t\end{array}$

Step 3. To find the ratio of kinetic energy to potential energy as:
$$\begin{gathered} r=\frac{K}{U} \\ \Rightarrow r=\frac{{\displaystyle \frac{1}{2}}m{\omega }^2{A}^2{\cos }^2\omega t}{{\displaystyle \frac{1}{2}}m{\omega }^2{A}^2{\sin }^2\omega t} \end{gathered}$$

Step 4. Upon simplifying, we get the value of the ratio as:

$r=\frac{{\cos }^2\omega t}{{\sin }^2\omega t}$

$\begin{array}{l}\Rightarrow r=co{t}^2\omega t\left(\because cot\theta =\frac{\cos \theta }{\sin \theta }\right)\\ \Rightarrow r=co{t}^2\frac{\pi }{90}\left(210\right)\\ \Rightarrow r=co{t}^2\frac{7\pi }{3}\\ \therefore r=\frac{1}{3}\end{array}$

Hence, the correct option is $\left(B\right)$