InterviewSolution
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InterviewSolution
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A particle moving in a straight line has velocity v given by v2 = α – βy2, where α and β are constant and y is its distance from a fixed point in the line. Show that the motion of the particle is S.H.M. Find its time period and amplitude. |
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Answer» Given that: v2 = α − βy2 …(i) Differentiating it with respect to time t, we have 2v\(\frac{dv}{dt}\) = −2βy\(\frac{dy}{dt}\) or \(\frac{dv}{dt}\) = −βy…(ii) It means acceleration a =\(\frac{dv}{dt}\)= −βy As a and y have –ve sign shows that acceleration is directed towards mean position, so, if the particle is left free, it will execute S.H.M. Here ω2 = β ω = \(\sqrt{\beta}\) ∴ Time period, T = \(\frac{2\pi}{\omega}\) = \(\frac{2\pi}{\sqrt \beta}\) We know that, v = 0, when y = r from (ii) 0 = α – βr2 or r = \(\sqrt{\frac{\alpha}{\beta}}\) i.e., amplitude, r = \(\sqrt{\frac{\alpha}{\beta}}\) |
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