question_answer

What is the velocity of the bob of pendulum at its mean position, if it is a1 to vertical height of 10 cm? \[(g=9.8\,m/{{s}^{2}})\]

A)  \[2.2\text{ }m/s\]          

B)  \[1.8\text{ }m/s\]          

C)         \[1.4\text{ }m/s\]       

D)         \[0.6\text{ }m/s\]

Correct Answer: C

Solution :

Key Idea: Total energy remains conserved in simple harmonic motion. The bob possess kinetic energy at its mean position which gets converted to potential energy at height\[h.\]But the total energy remains conserved.         Hence, we have                                 \[KE=PE\] Let velocity of bob at mean position to v and m be its mass, then we have                                 \[\frac{1}{2}m{{v}^{2}}=mgh\]                 \[\Rightarrow \]               \[v=\sqrt{2gh}\]                 Putting \[g=9.8m/{{s}^{2}},h=0.1\,m\]                 \[\therefore \]  \[v=\sqrt{2\times 9.8\times 0.1}=1.4\,m/s.\]