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

Step 1: Understanding the Concept:
This system behaves like a bifilar suspension or a special case of a simple pendulum.
When the mass is displaced "out of the plane" of the strings, it rotates or swings about the axis PQ.
The effective length of this oscillation is the perpendicular distance from the mass to the axis PQ.
Step 2: Key Formula or Approach:
Time period of a simple pendulum: \(T = 2\pi \sqrt{\frac{l_{\text{eff}}}{g}}\).
Find the vertical height \(h\) from the mass R to the line PQ using the geometry of triangle PRQ.
Step 3: Detailed Explanation:
Consider the triangle formed by points P, Q, and R.
It is an isosceles triangle with sides \(PR = QR = L\) and base \(PQ = 2d\).
Let M be the midpoint of PQ. Triangle PMR is a right-angled triangle where \(PM = d\) and \(PR = L\).
The height of this triangle, which is the perpendicular distance from R to PQ, is: \[ h = \sqrt{L^2 - d^2} = (L^2 - d^2)^{1/2} \] For small oscillations out of the plane, the mass swings along an arc of a circle centered at point M on axis PQ. The effective radius of this swing is \(h\).
Substituting \(l_{\text{eff}} = h\) into the time period formula: \[ T = 2\pi \sqrt{\frac{(L^2 - d^2)^{1/2}}{g}} \] Step 4: Final Answer:
The time period of oscillation is \(2\pi \sqrt{\frac{(L^2 - d^2)^{1/2}}{g}}\).