Question
Consider a spring-mass simple harmonic oscillator in one dimension. The mass of the particle is $m\text{ }kg$ and the spring constant is $k{Nm}^{- 1}$. At a given instant, the extension of the spring is $x$-meter and the speed of the particle is $v{\text{ }ms}^{- 1}$. On the $x - v$ plane, if the graph of $v$ as a function of $x$ is a circle, then the correct option is:
Options
Solution
Sol. [Key Concept: Spring-mass SHM — ω = √(k/m)]
Velocity $v$ as a function of $n$ is given by
$${\begin{matrix} & v = \omega\sqrt{A^{2} - x^{2}} \\ & v^{2} = \omega^{2}\left( A^{2} - x^{2} \right) \\ & \frac{v^{2}}{\omega^{2}} + x^{2} = A^{2} \end{matrix} } {Given\ that\ above\ equation\ is\ of\ a\ circle }{Hence,\ \omega^{2} = 1 } \begin{matrix} & \frac{k}{m} = 1 \\ & k = m \end{matrix}$$
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