Question

What is the velocity of the bob of a simple pendulum at its mean position, if it is able to rise to a maximum vertical height of $40\text{cm}$.
(Take $g=9.8{\text{m/s}}^2$)

• A. $1.4\text{m/s}$
• B. $0.7\text{m/s}$
• C. $2.8\text{m/s}$
• D. $3.6\text{m/s}$

Answer 1

The correct option is C $2.8\text{m/s}$

$\Rightarrow$Mean position refers to the bottom most position of the bob.


Applying energy conservation between initial to final position,
$\text{Loss in KE of bob}=\text{Gain in PE of bob}$
$\Rightarrow \frac{1}{2}m{v}^2=mgh$
Or, $v=\sqrt{2\times g\times h}$
$v=\sqrt{2\times 9.8\times 0.4}=\sqrt{\frac{784}{100}}$
$\therefore v=2.8\text{m/s}$