Question

The bob of a pendulum was released from a horizontal position. The length of the pendulum is 10 m. If it dissipates 10% of its initial energy against air resistance, the speed with which the bob arrives at the lowest point is: [Use, $g : 10 \, \text{m/s}^2$]

• A. $6\sqrt{5} \, \text{ms}^{-1}$
• B. $5\sqrt{6} \, \text{ms}^{-1}$
• C. $5\sqrt{5} \, \text{ms}^{-1}$
• D. $2\sqrt{5} \, \text{ms}^{-1}$

Answer 1

Answer: (A)

$6\sqrt{5} \, \text{ms}^{-1}$

Answer 2

Given:
- Length of the pendulum, $\ell = 10 \, \text{m}$
- Gravitational acceleration, $g = 10 \, \text{m/s}^2$
- 10% of the initial energy is dissipated against air resistance.

Step 1. Calculate the initial energy:
Since the bob is released from a horizontal position, the initial potential energy (at the top) is:

$\text{Initial energy} = mg\ell$

Step 2. Calculate the energy at the lowest point:
Since 10% of the initial energy is dissipated, only 90% of the initial energy is available at the lowest point.

$\text{Energy at the lowest point} = \frac{9}{10} mg\ell$

Step 3. Relate energy to speed at the lowest point:
At the lowest point, this energy is entirely kinetic, so:

$\frac{9}{10}mg\ell = \frac{1}{2}mv^2$

Simplify by canceling $m$ from both sides:

$\frac{9}{10} \times 10 \times 10 = \frac{1}{2}v^2$

$v^2 = 180$

$v = \sqrt{180} = 6\sqrt{5} \, \text{m/s}$

Thus, the speed of the bob at the lowest point is $6\sqrt{5} \, \text{m/s}$.

The Correct Answer is: $6\sqrt{5} \, \text{m/s}$