Question

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

• 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

Answer: (A)

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