Question

A simple pendulum is suspended from the ceiling of a lift. When the lift is at rest its time period is $T$. With what acceleration should the lift be accelerated upwards in order to reduce its period to $T/2$ ? (g is acceleration due to gravity).

• A. 2 g
• B. 3 g
• C. 4 g
• D. g

Answer 1

Answer: (B)

3 g

Answer 2

Time period of simple pendulum is given by
$T=2 \pi \sqrt{\frac{l}{g}} \,\,\,\,\,\,\,\,...(i)$
When the lift is moving up with an acceleration $a$, then time period becomes
$T'=2 \pi \sqrt{\frac{l}{g+a}}$
Here,$T'=\frac{T}{2}$
$\Rightarrow \,\,\,\frac{T}{2}=2 \pi \sqrt{\frac{l}{g+a}}\,\,\,\,\,\,\,\,...(ii)$
Dividing E (ii) by E (i), we get
$a=3\, g$