JEE PYQ: Oscillations Question 33
Question 33 - 2019 (09 Jan 2019 Shift 2)
A particle is executing simple harmonic motion (SHM) of amplitude $A$, along the $x$-axis, about $x = 0$. When its potential Energy (PE) equals kinetic energy (KE), the position of the particle will be:
(1) $\frac{A}{2}$
(2) $\frac{A}{2\sqrt{2}}$
(3) $\frac{A}{\sqrt{2}}$
(4) $A$
Show Answer
Answer: (3)
Solution
PE = KE gives $x^2 = A^2 - x^2$, so $x = \frac{A}{\sqrt{2}}$.
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