Describe the vertical oscillations of a spring.

Explanation

Vertical oscillations of a spring: Let us consider a massles 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 br L. If the block of mass m is attched to the other end of spring , then spring elongates by a length l. Let `F_(1)` be the restoring force due to streching of spring . Due to mass m, the gravitatinal force acts vertically downword . We can draw free-body diagram for this system as shown in figure . Whwn the systym is under equilibrium,
`F_(1)+mg=0`
but the spring elongates by small displacement I, therefore,
`F_(1)propl impliesF_(1)=-kl`
`-kl+mg=0`

`mg=l or (m)/(k)=(l)/(g)`
Suppose we apply a vey small external force on the mass such that the mass further displaces downward bt a displacement y, then it will oscillate up and down , Now the restoring force due to this streching of spring (totl extension of spring is y+1) 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 the free body diagram for this case , we get ,
`-ky-kl+mg=m(d^(2)y)/(dt^(2)`
The net force acting on the mass due to this streching is `F=F_(2)+mg`
`F=-ky-kl+mg`
The gravitational force oppses the restroing force. Substituing equation (3) in equation (6), we get `F=-ky-kl+kl=-ky`
The above equation is in the form of simple harmonic differntial equation. Therefore, we get the time period as `T=2pisqrt(m/k) second`
The time period can be rewritten using equation (3)
`T=2pisqrt(m/k)=2pil(1)/(g)` second
The acceleratrion due to gravity g can be compound from the formula. `g=4pi^(2)((l)/(T^(2))) ms^(-2)`