Question

The period of a simple pendulum measured by inside a stationary lift 'T' . If the lift starts moving upward with a acceleration g/3 . What will be the time period?

• A. T/3
• B. 3T
• C. $\frac{\sqrt{3}}{2}T$
• D. $\sqrt{\frac{3}{2}}T$

Answer 1

The correct option is C $\frac{\sqrt{3}}{2}T$

Time period of pendulum, $T_P=2\pi \sqrt{\frac{stringlength}{downwardacceleration}}=2\pi \sqrt{\frac{L}{a_y}}$

At stationary lift, acceleration is $a_y=g$

Time period is $T=2\pi \sqrt{\frac{L}{g}}....\left(1\right)$

Lift accelerate upward, $a=g+\frac{g}{3}=\frac{4g}{3}$

Time period is $T^{\prime }=2\pi \sqrt{\frac{L}{4g/3}}=\frac{\sqrt{3}}{2}T$

Hence, time period will be $\frac{\sqrt{3}}{2}T$