For what value of displacement the kinetic energy and potential energy of a simple harmonic oscillation become equal?

Question

For what value of displacement the kinetic energy and potential energy of a simple harmonic oscillation become equal? (A) x =0 (B) x=± A (C) x =± ( A /√2) (D) x =( A /2)

Tardigrade Admin · Accepted Answer

KE = PE (1/2) m ω2( A 2- x 2)=(1/2) m ω2 x 2 A 2- x 2= x 2 2 x 2= A 2 x =± ( A /√2)

Q. For what value of displacement the kinetic energy and potential energy of a simple harmonic oscillation become equal?

$x = 0$

$x = \pm A$

$x = \pm \frac{A}{2}$

$x = \frac{A}{2}$

$K E = PE$
$\frac{1}{2} m ω^{2} (A^{2} - x^{2}) = \frac{1}{2} m ω^{2} x^{2}$
$A^{2} - x^{2} = x^{2}$
$2 x^{2} = A^{2}$
$x = \pm \frac{A}{2}$

Q. For what value of displacement the kinetic energy and potential energy of a simple harmonic oscillation become equal?

x=0

x=±A

x=±2​A​

x=2A​

KE=PE 21​mω2(A2−x2)=21​mω2x2 A2−x2=x2 2x2=A2 x=±2​A​

$x = 0$

$x = \pm A$

$x = \pm \frac{A}{2}$

$x = \frac{A}{2}$

$K E = PE$
$\frac{1}{2} m ω^{2} (A^{2} - x^{2}) = \frac{1}{2} m ω^{2} x^{2}$
$A^{2} - x^{2} = x^{2}$
$2 x^{2} = A^{2}$
$x = \pm \frac{A}{2}$