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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 . The distance of the particle from the mean position when its speed becomes 5 cm/s isA. `sqrt(3)` cmB. `sqrt(5)` cmC. `2 sqrt(3)` cmD. `2 sqrt(5)` cm |
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Answer» Correct Answer - C `V_(max) = a omega` `rArr" "omega = (V_(max))/(a) = (10)/(4) rads^(-1)` Now, `v = omega sqrt(a^(2) - y^(2))` `rArr" "v^(2) = omega^(2) (a^(2)-y^(2))` `rArr" "y^(2) = a^(2) - (v^(2))/(omega^(2))` `rArr" "y = sqrt(a^(2) - (v^(2))/(omega^(2)))` `= sqrt(4^(2) - (5^(2))/((10//4)^(2)))=2 sqrt(3)` cm |
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