Verified Solution • 2025–26 Session moderate

QAIPMT-2009-Ph-2147: NEET 2009 Physics - Class 11 AIPMT-2009-Ph-2147 Solution: A simple pendulum performs simple harmonic motion about x = 0…

A simple pendulum performs simple harmonic motion about x = 0 with an amplitude a and time period T. The speed of the pendulum at x = a/2 will be πa3T\frac{\pi a\sqrt{3}}{T} πaT\frac{\pi a}{T} πa32T\frac{\pi a\sqrt{3}}{2T} πa3T\frac{\pi a\sqrt{3}}{T}

Multiple Choice Options

  • A Acceleration = −- _ k (x + a)
  • B Acceleration = k(x + a)
  • C Acceleration = −- _ ω\omega _ 2 x
  • D Acceleration = −- _ k 0 x + k 1 x 2

Step-by-Step Verified Solution

Simple harmonic motion is defined as follows Acceleration d 2 y d t 2 = _ _ 2 x The negative sign is very important in simple harmonic motion. Acceleration is independent of any initial displacement of equilibrium position. Then acceleration = _ _ 2 x .

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