A particle executes simple harmonic motion with an amplitude of 4 cm. At the mean position the velocity of the particle is 10 cm s⁻¹. The distance of the particle from the mean position when its speed becomes 5 cm s⁻¹ is √α cm, where α= .

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12. At the mean position vₘₐₓ=Aω, so ω=(10)/4=2.5 rad s⁻¹. v=ω√A²-x² gives 5=2.5√16-x², so 16-x²=4 and x²=12, that is α=12.

A particle executes simple harmonic motion with an amplitude of 4 cm. At the mean position the velocity of the particle is 10 cm s⁻¹. The distance of the particle from the mean position when its speed becomes 5 cm s⁻¹ is √α cm, where α= ______.

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