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Displacement between max. P.E. position and max. K.E. position for a particle executing simple harmonic motion is : -
A
$\pm\dfrac{a}{2}$
B
$+a$
C
$\pm a$
D
$-1$
Detailed Solution
In SHM of amplitude a, the kinetic energy is maximum at the mean position ($x = 0$), where the speed is maximum.
The potential energy is maximum at the extreme positions, $x = -a$ and $x = +a$.
[add image: extreme position x = $-a$ (PE max), mean position x = 0 (KE max), extreme position x = $+a$ (PE max)]
So the displacement from the position of maximum KE to a position of maximum PE is either $+a$ or $-a$.
So the displacement is $\pm a$.
The potential energy is maximum at the extreme positions, $x = -a$ and $x = +a$.
[add image: extreme position x = $-a$ (PE max), mean position x = 0 (KE max), extreme position x = $+a$ (PE max)]
So the displacement from the position of maximum KE to a position of maximum PE is either $+a$ or $-a$.
So the displacement is $\pm a$.