A particle executes simple harmonic motion with an amplitude of 4 cm. At the mean position, velocity of the particle is 10 cm/s. The distance of the particle from the mean position when its speed becomes 5 cm/s is √(α) cm, where α = ________. Numerical answer type. Work it out, then open the answer.

Answer: 12. vmax = Aω ⇒ 10 = 4ω ⇒ ω = 2.5 rad/s. v = ω√(A² - x²) ⇒ 5 = 2.5√(16 - x²). 16 - x² = 4 ⇒ x² = 12, so x = √(12) cm. So α = 12.

A particle executes simple harmonic motion with an amplitude of 4 cm… | JEE Main 2024 (27 Jan, Shift 1)

Q31JEE Main 2024 · 27 JAN · SHIFT 1Medium

Velocity & Acceleration in SHM

A particle executes simple harmonic motion with an amplitude of $4\text{ cm}$ . At the mean position, velocity of the particle is $10\text{ cm/s}$ . The distance of the particle from the mean position when its speed becomes $5\text{ cm/s}$ is $\sqrt{\alpha}\text{ cm}$ , where $\alpha =$ ________.

Numerical answer type. Work it out, then open the answer.

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Answer: 12

$v_{max} = A\omega \Rightarrow 10 = 4\omega \Rightarrow \omega = 2.5$ rad/s.
$v = \omega\sqrt{A^2 - x^2} \Rightarrow 5 = 2.5\sqrt{16 - x^2}$ .
$16 - x^2 = 4 \Rightarrow x^2 = 12$ , so $x = \sqrt{12}$ cm.
So $\alpha = 12$ .

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