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The amplitude of a particle executing SHM is 4 cm. At the mean position, the speed of the particle is 16 `cms^(-1)`. The distance of the particle from the mean position at which the speed of the particle becomes `8 sqrt(3) cms^(-1)` will beA. `2sqrt3` cmB. `sqrt3` cmC. 1 cmD. 2 cm |
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Answer» Correct Answer - D At mean position velocity is maximum i.e., `v_"max" = omega a rArr omega = v_"max"/a =16/4=4` `v=omega sqrt(a^2-y^2) rArr 8sqrt3 = 4sqrt(4^2-y^2)` `rArr 192=16(16-y^2) rArr 12=16-y^2 rArr y=2 cm ` |
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