Question 277: Oscillations

Question

A simple pendulum of length ' $l$ ' has a brass bob attached at its lower end. It's period is ' T '. A steel bob of the same size, having density ' $x$ ' times that of brass, replaces the brass bob. Its length is then so changed that the period becomes ' 2 T '. What is the new length?

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, $ l $ is the length of the pendulum, and $ g $ is the acceleration due to gravity.
For the pendulum with the initial brass bob, the period is:
$$ T = 2\pi \sqrt{\frac{l}{g}} $$
When the brass bob is replaced with a steel bob of the same size but $ x $ times the density, the material change does not directly affect the period because the period of a simple pendulum depends only on its length and gravity, not the mass or density of the bob.
However, you are then modifying the length such that the new period becomes $ 2T $:
$$ 2T = 2\pi \sqrt{\frac{l_{	ext{new}}}{g}} $$
Setting this equal to the original equation for $ T $ scaled by 2:
$$ 2 \left( 2\pi \sqrt{\frac{l}{g}} ight) = 2\pi \sqrt{\frac{l_{	ext{new}}}{g}} $$
Solving for $ l_{	ext{new}} $:
Simplify both sides:
$$ \sqrt{\frac{l_{	ext{new}}}{g}} = 2 \sqrt{\frac{l}{g}} $$
Square both sides to eliminate the square roots:
$$ \frac{l_{	ext{new}}}{g} = 4 \cdot \frac{l}{g} $$
Multiply through by $ g $:
$$ l_{	ext{new}} = 4l $$
The new length of the pendulum that makes the period $ 2T $ is $ l_{	ext{new}} = 4l $.
Therefore, the correct option is:
Option C: $ 4l $

Answer

$4 l x$

Answer

$\frac{4 l}{\mathrm{x}}$

Answer

$4 l$

Answer

$2 l$

Oscillations

A simple pendulum of length ' l ' has a brass bob attached at its lower end. It's period is ' T '. A steel bob of the same size, having density ' x ' times that of brass, replaces the brass bob. Its length is then so changed that the period becomes ' 2 T '. What is the new length?

Choose the correct answer: