Question 135: Oscillations

Question

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 :

Accepted Answer

Total energy of particle = $${1 \over 2}k{A^2}$$
Potential energy (v) = $${1 \over 2}$$ kx
2
Kinetic energy (K) = $${1 \over 2}$$ kA
2
$$-$$ $${1 \over 2}$$kx
2
According to the question,
Potential energy = Kinetic energy
$$ 	herefore $$  $${1 \over 2}$$kx
2
= $${1 \over 2}$$kA
2
$$-$$ $${1 \over 2}$$ kx
2
$$ \Rightarrow $$  kx
2
= $${1 \over 2}$$ kA
2
$$ \Rightarrow $$  x = $$ \pm $$ $${A \over {\sqrt 2 }}$$

Answer

$${A \over 2}$$

Answer

$${A \over {2\sqrt 2 }}$$

Answer

$${A \over {\sqrt 2 }}$$

Answer

A

Oscillations

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 :

Choose the correct answer: