Search any question & find its solution
Question:
Answered & Verified by Expert
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)
Options:
Solution:
1492 Upvotes
Verified Answer
The correct answer is:
$\frac{m g A^{2}}{2 L}$
$\mathrm{PE}=\frac{\mathrm{m} \omega^{2} \mathrm{x}^{2}}{2}$
$\omega=\sqrt{\frac{\mathrm{g}}{\mathrm{L}}}$, 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 $\mathrm{PE}=\frac{\mathrm{MgA}^{2}}{2 \mathrm{~L}}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.