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Question
At what distance from the mean position is the speed of a particle performing S.H.M. half its maximum speed. Given the path length of S.H.M. = 10 cm.
Solution
Given: v = `1/2`vmax, 2A = 10 cm
∴ a = 5 cm
v = ω`sqrt("A"^2-"x"^2)` and vmax = ωA
since c = `1/2`vmax,
ω`sqrt("A"^2-"x"^2)=(ω"A")/2`
∴ A2 − x2 = `"A"^2/4`
∴ x2 = `"A"^2-"A"^2/4=(3"A"^2)/4`
∴ x = ±`sqrt3/2`A = ± 0.866 × 5 = ± 4.33 cm
This gives the required displacement.
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