A mass $$m$$ is suspended from the two coupled springs connected in series. The force constant for springs are $${k_1}$$ and $${k_2}.$$ The time period of the suspended mass will be

Solution :
Derive an expression from the given values which must be similar to $$a = - {\omega ^2}x.$$   Then calculate time period from the values in place of $$\omega .$$

The situation is shown in figure. Consider two springs of spring constants $${k_1}$$ and $${k_2}.$$ Here, the body of weight $$mg$$  is suspended at the free end of the two springs in series combination. When the body is pulled downwards through a little distance $$y,$$ the two springs suffer different extensions say $${y_1}$$ and $${y_2}.$$ But the restoring force is same in each spring.
$$\eqalign{ & \therefore F = - {k_1}{y_1} \cr & {\text{and}}\,\,F = - {k_2}{y_2} \cr & {\text{or}}\,\,{y_1} = - \frac{F}{{{k_1}}} \cr & {\text{and}}\,\,{y_2} = - \frac{F}{{{k_2}}} \cr} $$
$$\therefore $$ Total extension, $$y = {y_1} + {y_2}$$
$$\eqalign{ & = - \frac{F}{{{k_1}}} - \frac{F}{{{k_2}}} \cr & = - F\left( {\frac{{{k_1} + {k_2}}}{{{k_1}{k_2}}}} \right) \cr & {\text{or}}\,\,F = - \left( {\frac{{{k_1}{k_2}}}{{{k_1} + {k_2}}}} \right)y \cr} $$
If $$k$$ is the spring constant of series combination of springs then $$F = - ky$$
$$\therefore k = \frac{{{k_1}{k_2}}}{{{k_1} + {k_2}}}$$
If the body is left free after pulling down, it will execute $$SHM$$  of period
$$\eqalign{ & T = 2\pi \sqrt {\frac{m}{k}} \cr & = 2\pi \sqrt {\frac{{m\left( {{k_1} + {k_2}} \right)}}{{{k_1}{k_2}}}} \cr} $$