InterviewSolution
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InterviewSolution
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A 5 kg collar is attached to a spring of spring constant 500`Nm^(-1)` . It slides without friction over a horizontal rod. The collar is displaced from its equilibrium position by 10 cm and released. Calculate (a) The period of oscillations (b) The maximum speed and (c) maximum acceleration of the collar. |
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Answer» (a) The periodm of oscillation as given by Eq. (14.21) is, `T = 2pi sqrt((m)/(k))= 2pi sqrt((5.0 kg)/("500 N m"^(-1))` `= (2pi//10)s ` `= 0.63 s` (b) the velocity of the collar executing executting SHM is given by `v (t)=-A omega sin (omega t + phi)` The maximum speed is given by, `v_(m) = A omega` `= 0.1 xx sqrt((k)/(m))` `= 0.1 xx sqrt(("500 N m"^(-1))/(5 kg))` `= 1 ms^(-1)` and it occurs at `x=0` (c) The acceleration of the collar at the displacement `x(t)` from the equilibrium is given by, `a (t) = -omega^(2) x(t)` `= - (k)/(m) x(t)` Therefore, the maximum acceleration is, `a_(max) = omega^(2)A` `= ("500 N m"^(-1))/("5 kg") xx 0.1 m` `= 10 ms^(-2)` and it occurs at the extremitles. |
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