Question 288: Oscillations

Question

The period of a simple pendulum gets doubled when

Accepted Answer

The period of a simple pendulum is given by the formula:
$$ T = 2\pi \sqrt{\frac{L}{g}} $$
where:
$ T $ is the period of the pendulum,
$ L $ is the length of the pendulum,
$ g $ is the acceleration due to gravity.
To determine when the period $ T $ gets doubled, consider the scenario where $ T' = 2T $. Using the period formula, we have:
$$ T' = 2\pi \sqrt{\frac{L'}{g}} = 2 \cdot 2\pi \sqrt{\frac{L}{g}} $$
This simplifies to:
$$ \sqrt{\frac{L'}{g}} = 2 \sqrt{\frac{L}{g}} $$
Squaring both sides gives:
$$ \frac{L'}{g} = 4 \cdot \frac{L}{g} $$
which implies:
$$ L' = 4L $$
Therefore, the period of a simple pendulum gets doubled when its length is made four times. The correct option is
Option B: its length is made four times
.

Answer

its length is doubled.

Answer

its length is made four times.

Answer

its length is made half.

Answer

the mass of the bob is doubled.

Oscillations

The period of a simple pendulum gets doubled when

Choose the correct answer: