JEE PYQ: Oscillations Question 20
Question 20 - 2021 (26 Feb 2021 Shift 2)
A particle executes S.H.M with amplitude ‘a’ and time period $T$. The displacement of the particle when its speed is half of maximum speed is $\frac{a\sqrt{x}}{2}$. The value of $x$ is ______
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Answer: 3
Solution
$\frac{a\omega}{2} = \omega\sqrt{a^2 - y^2}$. $y = \frac{a\sqrt{3}}{2}$. So $x = 3$.