Question

What is the velocity of the bob of pendulum at its mean position, if it is able to rise  to  a vertical height of \(10\, cm\) ? \((g=9.8\,m/s^{2})\)

• A. 2.2 m/s
• B. 1.8 m/s
• C. 1.4 m/s
• D. 0.6 m/s

Answer 1

The Correct Option is C

To find the velocity of the bob at its mean position, we can use the principle of conservation of mechanical energy. The potential energy (PE) at the maximum height is converted into kinetic energy (KE) at the mean position.
Step-by-Step Solution:|
1. Calculate the Potential Energy (PE) at the maximum height:
  The potential energy at the height \(h\) is given by:
  \[  PE = mgh  \]
  where:
  - \(m\) is the mass of the bob
  - \(g\) is the acceleration due to gravity (\(9.8\, \text{m/s}^2\))
  - \(h\) is the height (\(0.10\, \text{m}\))
2. Convert the Potential Energy (PE) into Kinetic Energy (KE) at the mean position:
  At the mean position, the height is zero, so all the potential energy will be converted into kinetic energy. The kinetic energy is given by:
  \[  KE = \frac{1}{2}mv^2  \]
  where:
  - \(v\) is the velocity of the bob at the mean position
3. Set PE equal to KE:
  \[ mgh = \frac{1}{2}mv^2\]
  Notice that the mass \(m\) cancels out from both sides of the equation:
  \[gh = \frac{1}{2}v^2\]
4. Solve for \(v\):
  \[ v^2 = 2gh \]
  \[v = \sqrt{2gh}\]
5. Substitute the known values:
  \[ v = \sqrt{2 \times 9.8\, \text{m/s}^2 \times 0.10\, \text{m}} \]
  \[v = \sqrt{1.96\, \text{m}^2/\text{s}^2}\]
  \[ v = 1.4\, \text{m/s} \]
The velocity of the bob at its mean position is \(1.4\, \text{m/s}\).