Vertical oscillations of a spring: Let us consider a massless spring with stiffness 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) prop l rArr F_(1) =` -kl ...(2)
Substituting equation (2) in equation (1), we get
-kl + mg = 0
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 + 1) = -ky – kl ...(4)
Since, the mass moves up and down with acceleration `(d^(2)y)/(dt^(2)` 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))` = -ky
`(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) = 2 pirl (l)/(g)` second ...(9)
The acceleration due to gravity g can be computed from the formula
`g = 4pi^(2) ((l)/(T^(2))) ms^(-2)` ...(10)
