Question

When a mass \(m\) is suspended from a spring of length \(\ell\), the length of the spring becomes \(L\). The mass is pulled down by a distance \(d\) and released. If the equation of motion of the mass is \( \frac{d^2x}{dt^2} + Px = 0 \), then \(P\) is equal to ( \(g\) = acceleration due to gravity )

• A. \( \dfrac{L-\ell}{g} \)
• B. \( \dfrac{g}{L-\ell} \)
• C. \( \sqrt{\dfrac{g}{L-\ell}} \)
• D. \( \sqrt{\dfrac{L-\ell}{g}} \)

Hint:

In spring–mass systems, always use the static equilibrium condition first to find the spring constant.

Answer 1

The Correct Option is C

Step 1: Extension of the spring.
The extension produced in the spring is \[ x = L - \ell. \]
Step 2: Using equilibrium condition.
At equilibrium, \[ mg = k(L - \ell), \] where \(k\) is the spring constant. Hence, \[ k = \frac{mg}{L - \ell}. \]
Step 3: Equation of motion.
For simple harmonic motion, \[ \frac{d^2x}{dt^2} + \frac{k}{m}x = 0. \] Comparing with the given equation, \[ P = \frac{k}{m} = \frac{g}{L - \ell}. \]
Step 4: Conclusion.
Thus, the value of \(P\) is \( \sqrt{\dfrac{g}{L-\ell}} \).