The angular velocity and the amplitude of a simple pendulum is $\omega$ and a respectively. At a displacement x from the mean position if its kinetic energy is T and potential energy is V, then the ratio of T to V is
- $ \frac{ ( a^2 - x^2 \, \omega^2 ) }{ x^2 \, \omega^2 }$
- $ \frac{ x^2 \, \omega^2}{ (a^2 - x^2 \, \omega^2 )}$
- $ \frac{ (a^2 - x^2) }{ x^2 } $
- $ \frac{ x^2 }{ (a^2 - x^2 )} $
The Correct Option is C
Solution and Explanation
and KE, T = $ \frac{1}{2} m \omega^2 (a^2 - x^2)$
$\therefore \frac{ T}{ V} = \frac{ a^2 - x^2 }{ x^2 } $
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Concepts Used:
Energy In Simple Harmonic Motion
We can note there involves a continuous interchange of potential and kinetic energy in a simple harmonic motion. The system that performs simple harmonic motion is called the harmonic oscillator.
Case 1: When the potential energy is zero, and the kinetic energy is a maximum at the equilibrium point where maximum displacement takes place.
Case 2: When the potential energy is maximum, and the kinetic energy is zero, at a maximum displacement point from the equilibrium point.
Case 3: The motion of the oscillating body has different values of potential and kinetic energy at other points.