Question

The time period of a simple pendulum is given by the equation $T=2\pi \sqrt{\frac{l}{g}}$. The equation of motion of the pendulum will be $\frac{d^{2}x}{d{t}^2}$ :

• A. $-\frac{l}{g}x$
• B. $\frac{g}{l}x$
• C. $\frac{l}{g}x$
• D. $-\frac{g}{l}x$

Answer 1

Answer: (D)

$-\frac{g}{l}x$

$T=2\pi \sqrt{\frac{l}{g}}$
Since $\omega =\frac{2\pi }{T}=\sqrt{\frac{g}{l}}$
Thus, ${\omega }^2=\frac{g}{l}$

and

Acceleration = $-{\omega }^{2}x=\frac{d^{2}x}{d{t}^2}=-\frac{gx}{l}$