Question

The angular velocity and the amplitude of simple pendulum is $\omega$ and $a$ respectively. At a displacement $X$ from mean position, if the kinetic energy is $T$ and potential energy is $V$. Then ratio $T$ to $V$

• A. $X^2{\omega }^2/(a^2-X^2{\omega }^2)$
• B. $X^2/(a^2-X^2)$
• C. $a^2-X^2{\omega }^2/X^2-X^2{\omega }^2$
• D. $(a^2-X^2)/X^2$

Answer 1

Answer: (D)

$(a^2-X^2)/X^2$

We know,

$T=\frac{1}{2}m{\omega }^2(a^2-X^2)$

$V=\frac{1}{2}m{\omega }^2{X}^2$

$T:V=\frac{a^2-X^2}{X^2}$

Option $\text{D}$ is the correct answer