Question

The time-period of a simple pendulum of length $\sqrt{5}$m suspended in a car moving with uniform acceleration of $5m{s}^{-2}$ in a horizontal straight road is $(g=10m{s}^{-2})$:

• A. $\frac{2\pi }{\sqrt{5}}s$
• B. $\frac{\pi }{\sqrt{5}}s$
• C. $5\pi s$
• D. $4\pi s$
• E. $3\pi s$

Answer 1

The correct option is A $\frac{2\pi }{\sqrt{5}}s$

Given, length of simple pendulum, $l=\sqrt{5}$m
Acceleration of car (a)$=5m{s}^{-2}$
Now $g_{effective}=\sqrt{a^2+g^2}$
$g_{effective}=\sqrt{(5{)}^2+(10{)}^2}$
$=\sqrt{125}=5\sqrt{5}$
$T=2\pi \sqrt{\frac{l}{g_{effective}}}$
$T=2\pi \sqrt{\frac{\sqrt{5}}{5\sqrt{5}}}$
$T=\frac{2\pi }{\sqrt{5}}s$.