Question

The time period of a simple pendulum measured inside a stationary lift is found to be T. If the lift starts accelerating upwards with an acceleration g/3, the time period is

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

Answer 1

Answer: (B)

$T \sqrt { 3 } / 2$

Answer 2

$T = 2 \pi \sqrt { \frac { l } { g } }$ and $T ^ { \prime } = 2 \pi \sqrt { \frac { l } { 4 g / 3 } }$

$\left[ A s g ^ { \prime } = g + a = g + \frac { g } { 3 } = \frac { 4 g } { 3 } \right.$ ]

∴ $T ^ { \prime } =$ $\frac { \sqrt { 3 } } { 2 } T$