Question

Consider a spring-mass simple harmonic oscillator in one dimension. The mass of the particle is \(m\) kg and the spring constant is \(k\) N m\(^{-1}\). At a given instant, the extension of the spring is \(x\) metre and the speed of the particle is \(v\) m s\(^{-1}\). On the \(x-v\) plane, if the graph of \(v\) as a function of \(x\) is a circle, then

• A. \(k=\sqrt{m}\)
• B. \(k=\dfrac{1}{m}\)
• C. \(k=m\)
• D. \(k=m^2\)

Hint:

The phase-space plot of SHM is generally an ellipse. It becomes a circle when the coefficients of \(x^2\) and \(v^2\) become equal.

Answer 1

The Correct Option is C

Concept: For SHM, \[ \frac12 kx^2+\frac12 mv^2 = E \] This represents an ellipse in the \(x-v\) plane.

Step 1: Write the equation in standard form.
\[ \frac{kx^2}{2E} + \frac{mv^2}{2E} = 1 \] or \[ \frac{x^2}{2E/k} + \frac{v^2}{2E/m} = 1 \]

Step 2: Condition for a circle.
For a circle, both denominators must be equal. \[ \frac{2E}{k} = \frac{2E}{m} \] \[ k=m \]

Step 3: Write the final answer.
\[ \boxed{k=m} \] Hence, \[ \boxed{\text{Option (C)}} \]