Question

A pendulum is made to hang from the ceiling of an elevator. It has period of $T$ sec [for small angles). The elevator is made to accelerate upwards with $10\, m / s ^{2}$. The period of the pendulum now will be (Take $g =10\, m / s ^{2}$ )

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

Answer 1

Answer: (C)

When the elevator is at rest, its time period is given by
$T=2 \pi \sqrt{\frac{I}{g}}=2 \pi \sqrt{\frac{I}{10}}$
When the elevator accelerates upwards,
its time period becomes.
$T=2 \pi \sqrt{\frac{I}{g+a}}=2 \pi \sqrt{\frac{I}{10+10}} $
$=2 \pi \sqrt{\frac{I}{20}}=2 \pi \sqrt{\frac{I}{10}} \times \frac{1}{\sqrt{2}}$
$=\frac{T}{\sqrt{2}}$