Describe the vertical oscillations of a spring.

Explanation

(i) Consider a massless spring with stiffness constant or force constant k attached to a ceiling.
(ii) 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 , the the spring elongates by a length l .
(iii) Let `F_(1)` be the restoration force due to stretching of spring . Due to mass m, 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)`
(iv) But the spring elongates by small displacement l , therefore
`F_(1) prop l implies F_(1) =-kl " "...(2)`
Substituting equation (2) in equation (1), we get
`-kl+mg=0`
`mg=kl`
or
`(m)/(k)=(l)/(g)" "....(3)`

(v) Suppose we apply a very small external force is applied 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 b(total extension of spring is y+l) is
`F_(2)prop (y+l) `
`F_(2)=-k(y+l)=-ky-kl`
Since, the mass moves up and down with acceleration `(d^(2)y)/(dt^(2))` by, drawing for this case , we get
`-ky-kl +mg=m (d^(2)y)/(dt^(2))`
(vi) The net force acting on the mass due to this stretching is
`F=F_(2) +mg`
`F=-ky-kl+mg" "....(4)`
The gravitational force oppose the restoring force. Substitute equation (3) in equation (4) , we get
`F=-ky-kl+kl=-ky`
Applying Newton's law , we get
`m(d^(2)y)/(dt^(2))=-ky,(d^(2)y)/(dt^(2))=-k/my`
(vii) The above equation is in the form of simple harmonic differential equation. Therefore we get the time period as
`T=2pisqrt((m)/(k))` second
The acceleration due to gravity g can be computed from the formula
(viii) The time period can be rewritten using equation
`T=2pisqrt((m)/(k))=2pisqrt((l)/(g))` second
The acceleration due to gravity g can be computed from the formula
`8=pi^(2)((l)/(T^(2)))ms^(-1)`