The angular velocity and the amplitude of a simple pendulum is ω and a respectively. At a displacement x from the mean position if its kinetic energy is T and potential energy is V, then the ratio of T to V is

Question

The angular velocity and the amplitude of a simple pendulum is ω and a respectively. At a displacement x from the mean position if its kinetic energy is T and potential energy is V, then the ratio of T to V is (A) (a2-x2 ω2) / x2 ω2 (B) x2 ω2 /(a2-x2 ω2) (C) (a2-x2) / x (D) x2 /(a2-x2)

Tardigrade Admin · Accepted Answer

Potential energy, V=(1/2) m ω2 x2 and kinetic energy E, T=(1/2) m ω2(a2-x2) ∴ (T/V)=(a2-x2/x2)

Q. The angular velocity and the amplitude of a simple pendulum is $ω$ and $a$ respectively. At a displacement $x$ from the mean position if its kinetic energy is $T$ and potential energy is $V$ , then the ratio of $T$ to $V$ is

$(a^{2} - x^{2} ω^{2}) / x^{2} ω^{2}$

$x^{2} ω^{2} / (a^{2} - x^{2} ω^{2})$

$(a^{2} - x^{2}) / x$

$x^{2} / (a^{2} - x^{2})$

Potential energy, $V = \frac{1}{2} m ω^{2} x^{2}$
and kinetic energy $E, T = \frac{1}{2} m ω^{2} (a^{2} - x^{2})$
$∴ \frac{T}{V} = \frac{a ^{2} - x ^{2}}{x ^{2}}$

Q. The angular velocity and the amplitude of a simple pendulum is ω and a respectively. At a displacement x from the mean position if its kinetic energy is T and potential energy is V, then the ratio of T to V is

(a2−x2ω2)/x2ω2

x2ω2/(a2−x2ω2)

(a2−x2)/x

x2/(a2−x2)