Question

‘P’ and ‘Q’ are fixed points in same plane and mass ‘m’ is tied by string as shown in figure. If the mass is displaced slightly out of this plane and released, it will oscillate with time period (\(PQ = 2d,\; PR = QR = L\))

• A. \(2\pi \sqrt{\frac{L}{g}}\)
• B. \(2\pi \sqrt{\frac{L^2}{g}}\)
• C. \(2\pi \sqrt{\frac{\sqrt{L^2 - d^2}}{g}}\)
• D. \(2\pi \sqrt{\frac{\sqrt{L^2 + d^2}}{g}}\)

Hint:

Effective length = perpendicular height of triangle.

Answer 1

The Correct Option is C

Concept:
For small oscillations perpendicular to plane, motion is SHM. Effective length is vertical height of triangle. Step 1: Geometry of system. Triangle \(PQR\) is isosceles: \[ PR = QR = L, \quad PQ = 2d \] Drop perpendicular from \(R\) to midpoint of \(PQ\): \[ \text{Height } h = \sqrt{L^2 - d^2} \]
Step 2: Physical interpretation. When displaced slightly out of plane, restoring force acts like simple pendulum of length \(h\).
Step 3: Time period formula. \[ T = 2\pi \sqrt{\frac{h}{g}} \]
Step 4: Substitute value. \[ T = 2\pi \sqrt{\frac{\sqrt{L^2 - d^2}}{g}} \]
Step 5: Conclusion. \[ T = 2\pi \sqrt{\frac{\sqrt{L^2 - d^2}}{g}} \]