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

Step 1: Write the energy equation.
For a spring-mass oscillator, total energy is constant: $\tfrac12 k x^2 + \tfrac12 m v^2 = E$. This links $x$ and $v$.
Step 2: Recognise the shape.
An equation of the form $(\text{constant})x^2 + (\text{constant})v^2 = E$ describes an ellipse in the $x$-$v$ plane.
Step 3: Put it in standard form.
Dividing by $E$: $\dfrac{x^2}{2E/k} + \dfrac{v^2}{2E/m} = 1$.
Step 4: State the circle condition.
An ellipse becomes a circle only when the two denominators are equal: $\dfrac{2E}{k} = \dfrac{2E}{m}$.
Step 5: Solve the condition.
Cancelling $2E$ gives $\dfrac1k = \dfrac1m$, hence $k = m$.
Step 6: Choose the option.
The graph is a circle when $k = m$, which is option C.
\[ \boxed{ k = m } \]