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