When the displacement of a particle executing simple harmonic motion is half its amplitude, the ratio of its kinetic energy to potential energy is

Question

When the displacement of a particle executing simple harmonic motion is half its amplitude, the ratio of its kinetic energy to potential energy is (A) 1: 3 (B) 2: 1 (C) 3: 1 (D) 1: 2

Tardigrade Admin · Accepted Answer

x=A cos ω t x=A / 2 text we get ω t=60° (K E/P E)=((1/2) m ω2 A2 sin 2 ω t/(1)2 m ω2 A2 cos ω t)= tan 2 ω t = tan 2 60°=3

Q. When the displacement of a particle executing simple harmonic motion is half its amplitude, the ratio of its kinetic energy to potential energy is

$1 : 3$

$2 : 1$

$3 : 1$

$1 : 2$

$x = A cos ω t$
$x = A /2$
$we get ω t = 6 0^{\circ}$
$\frac{K E}{PE} = \frac{\frac{1}{2} m ω ^{2} A ^{2} s i n ^{2} ω t}{\frac{1}{2} m ω ^{2} A ^{2} c o s ω t} = tan^{2} ω t$
$= tan^{2} 6 0^{\circ} = 3$