Figure `(i)` 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`(ii)` shows spring with both endsd free and attached to a mass `m` at either end. Each end of the spring in figure `(ii)` is stretched by the same force `F`.

(a) What is the maximum extension of the spring in the two cases ?
(b) If the mass in figure `(i)` and the two masses in Figure `(ii)` are released free. what is the period of oscillation in each case ?

Answer

(a) Maximum extersion `= (F)/(K)` (for both cases)
(b) in fig. (a), `m` is the inertia factor and `k` is the spring factor.
`T = 2pisqrt(("inertia factor")/("spring factor"))` or `T = 2pisqrt((m)/(k))`
in Fig. (b), `m(d^(2)x_(1))/(dt^(2))=-kx` and `m(d^(2)x_(2))/(dt^(2)) = kx`
Change in length of spring `x = (x_(1) - x_(2))`
or `(d^(2)x)/(dt^(2)) = (d^(2))/(dt^(2)) (x_(1)-x_(2))` [Note that `I` is constant]
or `m(d^(2)x)/(dt^(2)) + 2kx = 0`
or `(d^(2)x)/(dt^(2)) + (2k)/(m)x = 0` or `(d^(2)x)/(dt^(2)) + omega^(2) = 0`
or `omega = sqrt((2k)/(m)) rArr T = (2pi)/(omega) = 2pisqrt((m)/(2k))`
Alternate : Reduced mass of the given two-boy system.
`mu = (m xx m)/(m + m) = (m^(2))/(2m) = (m)/(2)`
inertia factor `= (m)/(2)` and spring factor `= k`
`:.` Time period , `T = 2pisqrt((m//2)/(k)) = 2pisqrt((m)/(2k))`