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PhysicsNEET 2024Spring–mass system and simple pendulum

Oscillations

If the mass of the bob in a simple pendulum is increased to thrice its original mass and its length is made half its original length, then the new time period of oscillation is x2\dfrac{\sqrt{x}}{2} times its original time period. Then the value of xx is:

  1. A.2✓
  2. B.232\sqrt{3}
  3. C.4
  4. D.3

Solution

T=2πlgT = 2\pi\sqrt{\frac{l}{g}} is independent of mass. Halving ll gives T′=T2=22TT' = \frac{T}{\sqrt2} = \frac{\sqrt2}{2}T, so x=2x = 2.

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