JEE MAIN - Physics (2019 - 9th January Evening Slot - No. 16)
A particle is executing simple harmonic motion (SHM) of amplitude A, along the x-axis, about x = 0. When its potential Energy (PE) equals kinetic energy (KE), the position of the particle will be :
$${A \over 2}$$
$${A \over {2\sqrt 2 }}$$
$${A \over {\sqrt 2 }}$$
A
Explanation
Total energy of particle = $${1 \over 2}k{A^2}$$
Potential energy (v) = $${1 \over 2}$$ kx2
Kinetic energy (K) = $${1 \over 2}$$ kA2 $$-$$ $${1 \over 2}$$kx2
According to the question,
Potential energy = Kinetic energy
$$ \therefore $$ $${1 \over 2}$$kx2 = $${1 \over 2}$$kA2 $$-$$ $${1 \over 2}$$ kx2
$$ \Rightarrow $$ kx2 = $${1 \over 2}$$ kA2
$$ \Rightarrow $$ x = $$ \pm $$ $${A \over {\sqrt 2 }}$$
Potential energy (v) = $${1 \over 2}$$ kx2
Kinetic energy (K) = $${1 \over 2}$$ kA2 $$-$$ $${1 \over 2}$$kx2
According to the question,
Potential energy = Kinetic energy
$$ \therefore $$ $${1 \over 2}$$kx2 = $${1 \over 2}$$kA2 $$-$$ $${1 \over 2}$$ kx2
$$ \Rightarrow $$ kx2 = $${1 \over 2}$$ kA2
$$ \Rightarrow $$ x = $$ \pm $$ $${A \over {\sqrt 2 }}$$
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