Question

The length of a simple pendulum is $32\text{cm}$. It is suspended by the roof of a lift which is moving upwards with an acceleration of $6.2{\text{m/s}}^2$. Find the time period of pendulum.

• A. $0.2\sqrt{2}\pi \text{s}$
• B. $\sqrt{2}\pi \text{s}$
• C. $0.6\sqrt{2}\pi \text{s}$
• D. $0.2\sqrt{6}\pi \text{s}$

Answer 1

The correct option is A $0.2\sqrt{2}\pi \text{s}$

Given,
Length of simple pendulum, $\left(l\right)=32\text{cm}$
Acceleration of the lift $\left(a\right)=6.2{\text{m/s}}^2$
Since, Time period $\left(T\right)=2\pi \sqrt{\frac{l}{g_{eff}}}$
Lift is moving upward with an acceleration $a$
$\therefore g_{eff}=g+a\Rightarrow T=2\pi \sqrt{\frac{l}{g+a}}=2\times \pi \times \sqrt{\frac{0.32}{9.8+6.2}}$
$=2\times \pi \times \sqrt{\frac{0.32}{16}}$
$=0.2\sqrt{2}\pi \text{s}$