Question:
Vertical displacement of a plank with a body of mass m on it is varying according to law
y = sin ω t + √(3)cos ω t.
The minimum value of ω for which the mass just breaks contact with the plank and the moment it occurs first after t = 0, are given by
Vertical displacement of a plank with a body of mass m on it is varying according to law
y = sin ω t + √(3)cos ω t.
The minimum value of ω for which the mass just breaks contact with the plank and the moment it occurs first after t = 0, are given by
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Loss of contact occurs when downward acceleration equals g.
Updated On: Mar 20, 2026
- \( \sqrt{g/2},\; \frac{\sqrt{2}\pi}{6\sqrt{g}} \)
- \( \sqrt{g/2},\; \frac{2\pi}{3\sqrt{g}} \)
- \( \sqrt{g/3},\; \frac{\pi}{3\sqrt{2/g}} \)
- √(2g), (2π)/(3g)
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The Correct Option is A
Solution and Explanation
Step 1: Resultant displacement: y = 2sin(ω t + (π)/(3))
Step 2: Maximum downward acceleration: amax = 2ω²
Step 3: For loss of contact: 2ω² = g ⟹ ω = √((g)/(2))
Step 4: First occurrence after t=0: ω t + (π)/(3) = (3π)/(2) ⟹ t = frac√(2)π6√(g)
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