Describe the vertical oscillations of a spring.

Explanation

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 to mass 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)`