Question

How does the energy of a simple pendulum vary as it moves from one extreme position to the other during its oscillations?

Answer 1

we know, Total energy associated with a particle executing simple harmonic motion, at any point is the sum of kinetic energy and potential energy at that point.

e.g., $T.E=K.E+P.E$

we know, Kinetic energy is given by $K.E=\frac{1}{2}m\omega^2(A^2-y^2)$

and potential energy is given by, $P.E=\frac{1}{2}m\omega^2y^2$

so, total energy, T.E = $\frac{1}{2}m\omega^2(A^2-y^2)+\frac{1}{2}m\omega^2y^2$

=$\frac{1}{2}m\omega^2A^2$

at mean position, y = 0, P.E = 0
and $K.E_{max}=\frac{1}{2}m\omega^2A^2$

at extreme position, y =A , K.E = 0
and $P.E_{max}=\frac{1}{2}m\omega^2A^2$

here it is clear that , from mean position to extreme position , kinetic energy is to be converted into potential energy.

Answer 2

HEYA MATE HERE U GO.

❤. At the extreme postion , Velocity of the simple pendulum (i.e., bob ) will be zero.So its KE will also be equal to zero.Its total energy will be in the form of P.E

❤. When the bob comes towards mean postiton , its velocity gradually increases and becomes maximun at the mean position .So kE also gradually increases and becomes maximum at the mean position.But its p.E decreases by the same amount so that T.E will be same at every point.

❤. As the bob goes from mean position to another extreme position , its velocity gradually decreases and becomes zero at the extreme position.So also kE At this time P.E of the bob increases but T.E of the bob always remain constant.