MHT CET · Physics · Oscillations
A simple pendulum of length 'L' has mass 'm' and it oscillates freely with amplitude 'A' At extreme position, its potential energy is (g =acceleration due to gravity)
- A \(\frac{m g A^{2}}{2 L}\)
- B \(\frac{m g A^{2}}{L}\)
- C \(\frac{\operatorname{mgA}}{L}\)
- D \(\frac{m g A}{2 L}\)
Answer & Solution
Correct Answer
(A) \(\frac{m g A^{2}}{2 L}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{PE}=\frac{\mathrm{m} \omega^{2} \mathrm{x}^{2}}{2}\)
\(\omega=\sqrt{\frac{\mathrm{g}}{\mathrm{L}}}\), at extreame point \(\mathrm{x}=\mathrm{A}\)
So \(\mathrm{PE}=\frac{\mathrm{MgA}^{2}}{2 \mathrm{~L}}\)
\(\omega=\sqrt{\frac{\mathrm{g}}{\mathrm{L}}}\), at extreame point \(\mathrm{x}=\mathrm{A}\)
So \(\mathrm{PE}=\frac{\mathrm{MgA}^{2}}{2 \mathrm{~L}}\)
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