Question

Time period of a simple pendulum in a stationary lift is '$T$'. If the lift accelerates with $\frac{ g }{6}$ vertically upwards then the time period will be : (where $g =$ acceleration due to gravity)

• A. $\sqrt{\frac{6}{5}} T$
• B. $\sqrt{\frac{5}{6}} T$
• C. $\sqrt{\frac{6}{7}} T$
• D. $\sqrt{\frac{7}{6}} T$

Answer 1

Answer: (C)

$T =2 \pi \sqrt{\frac{\ell}{ g _{ eff }}}$

(a) when $a =0, T =2 \pi \sqrt{\frac{\ell}{ g }}$
(b) when $a=\frac{g}{6}, T^{\prime}=2 \pi \sqrt{\frac{\ell}{g+\frac{g}{6}}}$
$\therefore T ^{\prime}=\sqrt{\frac{6}{7}} T$