A pendulum bob is released from angular position \[\theta ,\]such that the magnitude of its initial acceleration and acceleration at lowest position are equal. |
The angular position \[\theta \] is |
A) \[\theta ={{\cos }^{-1}}\left( \frac{3}{5} \right)\]
B) \[\theta =si{{n}^{-1}}\left( \frac{3}{5} \right)\]
C) \[\theta ={{\tan }^{-1}}\left( \frac{3}{5} \right)\]
D) \[\theta ={{\cos }^{-1}}\left( \frac{2}{5} \right)\]
Correct Answer: A
Solution :
Let \[{{a}_{1}}\]and \[{{a}_{2}}\]are the accelerations, at given positions. Then, |
\[{{a}_{2}}=\frac{{{v}^{2}}}{l}=\frac{2gl\,(1-\cos \theta )}{l}\] |
Also, \[{{a}_{1}}=g\,\sin \theta \] |
As, \[{{a}_{1}}={{a}_{2}}\] |
\[\Rightarrow \]\[g\sin \theta =2g\,(1-\cos \theta )\] |
\[\Rightarrow \]\[\sin \theta =2\,(1-\cos \theta )\] |
\[\Rightarrow \]\[5{{\cos }^{2}}\theta -8\cos \theta +3=0\] |
\[\Rightarrow \]\[\cos \theta =\frac{3}{5}\]or \[\theta ={{\cos }^{-1}}\left( \frac{3}{5} \right)\] |
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