Let `x_0` be the maximum compression ( or extension) in the spring. In such situation

`K = 1/2 m (0)^2 = , U = 1/2 k x_0^2`

and `E = K +U = 1/2 kx_0^2 ..... (1)`

Now, if `v_0` be the maximum speed, of the block, when at mean position, then
`K = 1/2 m (v_0)^2 = , U = 1/2 k (0) =0`
`E = K+ U = 1/2 m (v_0)^2............ (2)`

But, as the block moves from its mean position, to either of its extreme position, it utilizes, its kinetic energy `(1/2 mv_0^2)` to elongate or compress; the spring, which gets stored in the spring in the form of potential energy`(1/2 k x_0^2)`.
Thus, from (1) and (2), it is evident at the total energy of the system (i.e., spring + block) remains constant.

`text(Note :)`

Oscillation of a spring pendulum is independent of the value of `g` at the place

Horizontal oscillations, of a block attached to a spring, on a rough surface.

Consider a block of mass 'm', attached to a spring of force constant k, oscillation on a rough horizontal surface with
coefficient of friction `mu`.

Let `x_0` be the initial compression, of the spring and `x_1` be the maximum extension, for the first time. From energy consideration

`1/2 kx_0^2 =1/2 kx_1^2 =(mumg)(x_0+x_1)`

`k/2(x_0+x_1)(x_0-x_1) = mu mg(x_0+x_1) => x_0-x_1 = (2 mu mg)/k`

Consider a block of mass `m` attached to one of an ideal spring, with its other end fixed to a rigid support. Let `y_0` be the initial extension in the spring. (Figure)

If 'k' be the force constant of the spring, then for equilibrium (vertical) of the block.

`ky_0 = mg` (numerically)

Now, if the block be slightly displaced from its mean position it starts executing SHM.
Let 'y' be the displacement of the block, at any instant from its mean position (downwards). The net force acting on the block upward will be

`F= k (y_0 + y) - mg`
or `vecF =- kvecy`
or acceleration `veca = vecF/m = -(k/m)vec y`

Comparing the above equation with the standard from `vecF =-omega^2 vecy` it is evident that the block executes SHM, with an angular frequency `omega = sqrt(k/m)`
The period of oscillation `T = (2 pi)/omega = 2pi sqrt(m/k)`

`text(Note :)` Even the vertical spring oscillation is also independent of the value of 'g' at that place.

As the block oscillates, so does the spring. However the velocity at each point of the spring is not the same. In the elongated (or compressed) state, the displacement of each section can be assumed to be linearly proportional from their respective distance from the fixed end.

Thus, if 'y' be the displacement of the block (i.e., the end of the spring connected to the block) at any instant, then the displacement of a
section distance 'x' from the fixed end will be `(yx)/L` where L is the length of the spring; similarity if v be
the velocity of block then `v = (dy)/(dt)` (Figure)

Accordingly the velocity `v` of the aforementioned section will be

`v = x/L` `(dy)/(dt)` `= (xv)/L`

Now, kinetic energy of the entire spring can be obtained as the sum (integral) of the kinetic energies of each section.

Thus, total K.E. of the spring `= int{1/2((M^'dx)/L)v^2}`

[Here M' is the mass of the spring and dx, the length of the elemental section.]

`= 1/2 (M^')/L int_0^L(v^2x^2dx)/L^2 = 1/2 (M^'v^2)/L^3[x^3/3]_0^L = 1/2 ((M^'v^2)/3)`

Total kinetic energy of the spring + block system `= 1/2 ((M^'v^2)/3) +1/2Mv^2` [where M is the mass of the block]

Now, the total mechanical energy of the system is given by `E = K + U`

`= 1/2 ((M^'v^2)/3)+ 1/2Mv^2 +1/2ky^2`

Since the total mechanical energy should remain conserved

`(dE)/(dt) = 0`

`(M^')/6 ((2vdv)/dt) + M/2((2vdv)/dt) +k/2(2y (dy)/(dt)) = 0`

`(M^'/3+M)(dv)/(dt) +ky =0`, [since `(dy)/(dt) =v`]

Again since `(dv)/(dt) = d/(dt)((dy)/(dt)) = (d^2y)/(dt^2)`

`(d^2y)/(dt)+ y/((M^')/3+M) =0`

Comparing with the standard form `(d^2y)/(dx^2) + omega^2 y = 0`

We find, that the complete system executes SHM, with an angular frequency `omega = sqrt(k/((M^')/3+M))`

Time period of oscillations `T = 2 pisqrt(((M^')/3+M)/k)`

`text(Note :)`

(1) The above equation can be rewritten as `T =2 pi sqrt(M_(eff)/k)` where `M_(eff)=((M^')/3+M )`

(2) Tile oscillations of a block attached to a spring, remains SHM only for� small elongation (or compression) when the restoring forces
generated in the spring remains linear. For large deformations in spring, the restoring forces become nonlinear.

(3) Unless, otherwise specified, a spring should always be treated as ideal and massless.

`(a)text(When springs are connected in series):`

Consider three springs with stiffness constants `k_1,k_2` and `k_3` respectively,

If `x_1, x_2` and `x_3` be the respective elongations of the three springs, then, the displacement of the block `x` is given by

`x= x_1 + x_2 + x_3 ....... (3)`

Now, if F be the restoring force acting along each of the spring

Then `F = k_1x_1 = k_2x_2 = k_3x_3 ........ (4)`

Now, the net force acting on the block will also be `F`
From `(4) x_1 = F/k_1, x_2 = F/k_2` and `x_3 = F/k_3`

Substituting them in equation (3)

`x = F/k_1 +F/k_2 +F/k_3`

or `(F/k_1 +F/k_2 +F/k_3) = F/(k_(eff))`

`1/k_(eff) = 1/k_1 +1/k_2 +1/k_3`

If m be the mass of the block, then acceleration

`veca = vecF/m = (-k_(eff)vecx)/m`

we have `veca = -(k_(eff)(vecx))/m`

which proves the fact that the oscillations are SHM, with angular frequency `omega = sqrt(k_(eff)/m)`
Time period of oscillation `= T = (2 pi)/omega => 2 pisqrt(m/(k_(eff)))`

where `1/k_(eff) = 1/k_1 +1/k_2 +1/k_3`


`text(When springs are connected in parallel :)`

It can be shown that the oscillations of the block are SHM with a time period T given by

`T = 2 pisqrt(m/k_(eff))`

where `k_(eff) = k_1 +k_2 +k_3`


`text(Figure-a shows a special arrangement of two springs virtually appearing to be in series; however they are not.)`

If `vecx` be the displacement of the block at any instant then, the net force acting on it

`= F(k_1 + k_2) `

The effective spring constant for the above set up is `k_(eff)= k_1 + k_2`

Time period `T = 2pi sqrt(m/k_(eff))`