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A body of mass `m` was suspended by a non-stretched spring, and then set fre without push. The stiffness of the spring is `x`. Neglecting the mass of the spring , find `:` `(a)` the law of motion `y(t)`, whee `y` is displacement of the body from the equilibrium position, `(b)` the maximum and minimum tensions of the spring in the process of motion. |
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Answer» Let `y(t)=` displacement of the body from the end of the unstretched position of the spring (not the equilibrium position). Then `m ddot(y)=-ky+mg` This equation has the soloution of the form `y=A+B cos (omegat+alpha)` if `- m omega^(2)B cos ( omegat +alpha)=-k[A+B cos(omegat+ alpha)]+ mg` Then ` omega^(2)=(k)/(m)` and `A=(mg)/(k)` we have `y=0` and `y=0` at `t=0`. So `- omega B sin alpha=0` ` A+Bcos alpha=0` Since `B gt 0` and `A gt 0` we have `alpha=pi` `B=A=(mg)/(k)` and `y=(mg)/(k)(1-cos omegat)` `(b)` Tension in the spring is `T=ky=mg(1-cos omegat)` so `T_(max)=2 mg, T_(m i n)=0` |
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