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) / x2 )  (D)  ( x2 / (a2 - x2 ))

Tardigrade Admin · Accepted Answer

P.E, V=(1/2) m ω2 x2 and KE , 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

$\frac{( a ^{2} - x ^{2} ω ^{2} )}{x ^{2} ω ^{2}}$

$\frac{x ^{2} ω ^{2}}{( a ^{2} - x ^{2} ω ^{2} )}$

$\frac{( a ^{2} - x ^{2} )}{x ^{2}}$

$\frac{x ^{2}}{( a ^{2} - x ^{2} )}$

P.E, $V = \frac{1}{2} m ω^{2} x^{2}$
and $K E, T = \frac{1}{2} m ω^{2} (a^{2} - x^{2})$
$∴ \frac{T}{V} = \frac{a ^{2} - x ^{2}}{x ^{2}}$

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

x2ω2(a2−x2ω2)​

(a2−x2ω2)x2ω2​

x2(a2−x2)​

(a2−x2)x2​