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Describe the vertical oscillations of a spring. |
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Answer» Solution :Vertical oscillations of a spring: Let us consider a massless spring with stiff NESS constant or force constant k attached to a ceiling as shown in figure. Let the LENGTH of the spring before loading mass m be L. If the block of mass m is attached to the other end of spring, then the spring ELONGATES by a length l. Let `F_(1)` be the restoring force due to stretching of spring. Due tomass m, the gravitational force acts vertically downward. We can draw free-body diagram for this system as shown in figure. When the system is under equilibrium, `F_(1)+mg=0""...(1)` But the spring elongates by small displacement l, therefore, `F_(1)proplimpliesF_(1)=-KL""...(2)` Substituting equation (2) in equation (1), we get `-kl+mg=0""...(3)` `mg=kl" (or) "(m)/(k)=(l)/(g)""...(3)` Suppose we apply a very small external force on the mass such that the mass further displaces downward by a displacement y, then it will oscillate up and down. Now, the restoring force due to this stretching of spring (total extension of spring is `y+1`) is `F_(2)prop(y+l)` `F_(2)=-k(y+l)=-ky-kl""...(4)` Since, the mass moves up and down with acceleration `(d^(2)y)/(dt^(2))`, by drawing the free body diagram for this case, we get `-ky-kl+mg=m(d^(2)y)/(dt^(2))""...(5)` The net force acting on the mass due to this stretching is `F=F_(2)+mg` `F=-ky-kl+mg""...(6)` The gravitational force opposes the restoring force. Substituting equation (3) in equation (6), we get `F=-ky-kl+kl=-ky` Applying Newton.s law, we get `m(d^(2)y)/(dt^(2))=-kyimplies (d^(2)y)/(dt^(2))=-(k)/(m)y"" ...(7)` The above equation is in the form of simple harmonic differential equation. Therefore, we get the time period as `T=2pisqrt((m)/(k))" second"...(8)` The time period can be rewritten using equation (3) `T=2pisqrt((m)/(k))=2pisqrt((l)/(g))""...(9)` The acceleration due to gravity g can be computed from the formula `g=4pi^(2)((1)/(T^(2)))ms^(-2)""...(10)` 
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