Simple Harmonic Motion 1 Question 4

4. For a simple pendulum, a graph is plotted between its Kinetic Energy (KE) and Potential Energy (PE) against its displacement $d$. Which one of the following represents these correctly? (graphs are schematic and not drawn to scale)

(2015 Main)

(a)

(c)

(b)

(d)

Show Answer

Answer:

Correct Answer: 4. (a)

Solution:

  1. Taking minimum potential energy at mean position to be zero, the expression of $KE$ and $PE$ are

$KE=\frac{1}{2} m \omega^{2}\left(A^{2}-d^{2}\right)$ and $PE=\frac{1}{2} m \omega^{2} d^{2}$

Both graphs are parabola. At $d=0$, the mean position,

$PE=0$ and $KE=\frac{1}{2} m \omega^{2} A^{2}=$ maximum

At $d= \pm A$, the extreme positions,

$KE=0$ and $PE=\frac{1}{2} m \omega^{2} A^{2}=$ maximum

Therefore, the correct graph is (a).



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