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InterviewSolution
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A spring of force constant `1200Nm^(-1)` is mounted on a horizontal table as shown in figure. A mass of 3.0kg is attached to the free end of the spring, pulled side ways to a distance of 2.0cm and released. Determing. (a) the frequency of oscillation of the mass. (b) the maximum acceleration of the mass. (c) the maximum speed of the mass. |
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Answer» `K = 1200 N m^(-1) ,m = 3 kg A = 2cm = 0.02m` (a) Frequecny, `f = (1)/(2pi) sqrt((K)/(m)) = (1)/(6.28) sqrt((1200)/(3)) = 3.2 Hz` (b) Acceleration `a = omega^(2) y = (K)/(m)y` Acceleration will be maximum when `y` is maximum i.e. `y=A` `:.` Max.acceleration, `a_(max) = (KA)/(m) = (1200 xx 0.02)/(3) = 8ms^(-2)` (c) Maximum speed of the mass will be when it is passing through the mean position, given by `V_(max) = A omega =A sqrt((K)/(m)) = 0.02 xx sqrt((1200)/(3)) = 0.4 ms^(-1)` |
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