Looking for classes? Ksquare Career Institute, Bengaluru →
Consider a spring-mass simple harmonic oscillator in one dimension. The mass of the particle is m 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 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:
Detailed Solution
Total energy $E=\frac{1}{2}mv^2+\frac{1}{2}kx^2 \Rightarrow \frac{v^2}{2E/m}+\frac{x^2}{2E/k}=1$. Comparing with $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, the curve is a circle when $a^2=b^2 \Rightarrow \frac{2E}{m}=\frac{2E}{k} \Rightarrow k=m$.