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 \[g=9.8m/{{s}^{2}},h=0.1\,m\] \[\therefore \] \[v=\sqrt{2\times 9.8\times 0.1}=1.4\,m/s.\]You need to login to perform this action.
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