Consider a spring-mass simple harmonic oscillator in one dimension. The mass of the particle is $m \mathrm{~kg}$ and the spring constant is $k \mathrm{Nm}^{-1}$. At a given instant, the extension of the spring is $x$-meter and the speed of the particle is $v \mathrm{~ms}^{-1}$. On the $x-v$ plane, if the graph of $v$ as a function of $x$ is a circle, then the correct option is:

For a simple pendulum, having time period $T$, the variation of kinetic energy (K.E.) with time $(t)$ is represented by:

The sum of kinetic energy and potential energy of a simple pendulum bob is 0.02 joule. The speed of the simple pendulum bob at equilibrium position is approximately:

(Consider mass of the bob $=20 \mathrm{~g}$ )

Savitha, a XI standard student, while conducting an experiment to determine the effective length of a simple pendulum $L$, notes down the data of time taken to complete 30 oscillations as 60 s and hence calculates the length of the simple pendulum as: (Take $\pi^2=9.8$, and $g=9.8 \mathrm{~m} / \mathrm{s}^2$ )