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MHT CET 202621 AprilEvening Shift
Exam
MHT CET
Year
2026
Date
21 April
Shift
Evening Shift
Subject
Physics
Chapter
Oscillations

MHT CET 2026 · 21 April · Evening Shift · Physics · Oscillations

If the kinetic energy of simple pendulum is (1)/(4) of its total...

If the kinetic energy of simple pendulum is 14\dfrac{1}{4} of its total energy, then the displacement (xx) and amplitude (AA) are related as

Options

  1. Ax=Ax = A
  2. Bx=2Ax = 2A
  3. Cx=32Ax = \dfrac{\sqrt{3}}{2}A
  4. Dx=34Ax = \dfrac{\sqrt{3}}{4}A

Correct answer

C. x=32Ax = \dfrac{\sqrt{3}}{2}A

Step-by-step solution

Total energy of a simple pendulum executing SHM is given by E=12mω2A2E = \dfrac{1}{2} m \omega^2 A^2

Kinetic energy at displacement xx is given by K=12mω2(A2−x2)K = \dfrac{1}{2} m \omega^2 (A^2 - x^2)

Given that K=14EK = \dfrac{1}{4} E

12mω2(A2−x2)=14(12mω2A2)\dfrac{1}{2} m \omega^2 (A^2 - x^2) = \dfrac{1}{4} \left( \dfrac{1}{2} m \omega^2 A^2 \right)

A2−x2=A24A^2 - x^2 = \dfrac{A^2}{4}

x2=A2−A24=3A24x^2 = A^2 - \dfrac{A^2}{4} = \dfrac{3A^2}{4}

x=32Ax = \dfrac{\sqrt{3}}{2} A

Answer: x=32Ax = \dfrac{\sqrt{3}}{2}A
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