For a simple pendulum, a graph is plotted between its kinetic energy (K.E.) and potential energy (P.E.) against its displacement d. which one of the following represents these correctly? (graphs are schematic and not

Question

Quantrex Academy · Accepted Answer

Correct answer: C.

For simple pendulum performing the simple harmonic motion, displacement,&

\text{y} = Asin\text{\omega t}v = \dfrac{dy}{dt} = \text{\omega A} cos \text{\omega t} = \text{\omega}\sqrt{A^{2} - y^{2}}\text{⇒K.E}\text{.} = \dfrac{1}{2}{mv}^{2}

,~
This represents a downward parabola,&

Y = \pm A

,~

\text{K.E.} = \dfrac{1}{2}{ω}^{2}\text{m}\left(A^{2} - A^{2}\right) = 0

.~

\text{P.E.} = \dfrac{1}{2}{mω}^{2}y^{2}

. This represents an upward parabola. On plotting graphs of potential energy & kinetic energy.
For a simple pendulum, a graph is plotted between its kinetic energy (K.E.) and potential energy (P.E.) against its

For a simple pendulum, a graph is plotted between its kinetic energy (K.E.) and potential energy (P.E.) against its

For a simple pendulum, a graph is plotted between its kinetic...

Options

Correct answer

Step-by-step solution

More from Oscillations