Figure 14.26 (a) shows a spring of force constant k clamped rigidly at one end and a mass m attached to its free end. A force F applied at the free end stretches the spring. Figure 14.26 (b) shows the same spring with both ends free and attached to a mass m at either end. Each end of the spring in Fig. 14.26(b) is stretched by the same force F.

What is the maximum extension of the spring in the two cases ?

Explanation

CASE (1) : Suppose (a) is the acceleration of mass m due to force F and extension produced in the spring is `y_1`.
From the Newton.s second law F= ma
Restoring force `F= -ky_(1) implies y_(1) = (F)/(k)` is the maximum extension of the spring now `ma = -ky_(1)`.
`therefore a= -(k)/(m)y_(1)`
Now acceleration in SHM is `a= -omega^(2)y`
`therefore omega^(2) = (k)/(m)`
`therefore omega =sqrt((k)/(m))" ""....."(1)`
Period of oscillation of spring, `T= (2pi)/(omega)`
`T= 2pi sqrt((m)/(k))" "(therefore" From equn. "(1))`.
CASE (2) : According to figure (b) applying force F on both body, the extension fo spring by `y_(2)` and extension of spring be y. on every body
`y_(2) = y.+y. = 2y.`
`therefore F= ma" and "F= -k(2y.) implies y. = (F)/(2k)`
is the extension of the spring now `ma = -2ky.`
`therefore a= -(k)/(m) (2y.)`, comparing with `a= -omega^(2)y`
`omega = sqrt((2k)/(m))`
`therefore (2pi)/(T) = sqrt((2k)/(m))`
`therefore T= 2pi sqrt((m)/(2k))`.