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
(A) A/2
(B) A/(2√2)
(C) A/√2
(D) A

Answer 1

Thus the position of the particle will be x = ±A / √2

Option (C) is correct.

Explanation:

Kinetic energy in S.H.M. -

K.E.= 1 / 2mu^2  

K.E = 1 / 2 m ( A^2 -x^2 )ω^2

K.E = mω^2

U = 1/2 Kx^2

K.E = 1/2 k ( A^2 -x^2 )

U = k

x = ±A / √2

Potential energy in S.H.M

P.E = 1/2 kx^2

where k = mω^2

Thus the position of the particle will be x = ±A / √2

Answer 2

The position of the particle When its potential energy (PE) equals kinetic energy (KE)  will be A/√2

total energy =1/2 kA²

at any point x, P.E =1/2kx²

and K.E=Total E -P.E

=1/2kA²-1/2kx²

when P.E=K.E we get

1/2kx² = 1/2kA²-1/2kx²

kx²=1/2kA²

x²=A²/2

⇒x =±A/√2