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The velocity v and displacement x of a particle executing simple harmonic motion are related as `v (dv)/(dx)= -omega^2 x`. `At x=0, v=v_0.` Find the velocity v when the displacement becomes x.A. `sqrt(v_(0)^(2)+omega^(2)x^(2))`B. `sqrt(v_(0)^(2)-omega^(2)x^(2))`C. `v =root(3)(v_(0)^(3)+omega^(2)x^(3))`D. `v=v_(0)-(omega^(3)x^(3)e^(x^(3)))^(1//3)` |
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Answer» Correct Answer - B (b) Given, `v(dv)/(dx)=-omega^(2)x` On integrating within the limit` underset(v_(0))overset(v)intvdv=underset(0)overset(x)-omega^(2)xdx` `implies[(v^(2))/(2)]_(v_(0))^(v)=-omega^(2)[(x^(2))/(2)]_(0)^(x)` `implies v^(2)-v_(0)^(2)=-omega^(2)x^(2)impliesv=sqrt(v_(0)^(2)-omega^(2)x^(2))` |
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