Explain the horizontal oscillations of a spring.

Explanation

Consider a massless spring with stiffness constant 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 length l. Let `F_(1)` be the restoring force due to stretching of spring. Due to mass m the gravitational force acts vertically downward. A free-body diagram is drawn for this system as shown in figure. When the system is under equilibrium,

`F_(1)+mg=0" ".......(1)`
But the spring elongets by small displacement l,
`:. F_(1) prop l rArr F_(1)=- kl " "....(2)`
Substituting eqaution (2) in equation (1) we, get
`-kg+mg=0`
`mg=kl`
(or) `(m)/(k)=(l)/(g)" "...(2)`
Substituting equation (2) in eqaution (1), we get
`-kl+mg=0`
`mg=kl`
Suppose a very 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 (total extension of spring is `y+l`) 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 of 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 gravvitational force oppose the resotoring force. Substituting eqution (3) in equation (6), we get
`F=-ky-kl+kl=-ky`
Time period :
Applying Newton's law, we get
`m(d^(2)y)/(dt^(2))=-ky`
`(d^(2)y)/(dt^(2))=-(k)/(m)y " "...(7)`
The above equation is in the form of simle harmonic differnetial equation. Hence the time period is
`T=2pi sqrt((m)/(k)) "second" " "...(8)`
The time period can be rewritten using equation (3) as
`T= 2pi sqrt((m)/(k))= 2pi sqrt((l)/(g))` second.