A block of mass `m` is tied to one end of a spring which passes over a smooth fixed pulley `A` and under a light smooth movable pulley `B`. The other end of the string is attached to the lower end of a spring of spring constant `K_2`. Find the period of small oscillation of mass `m` about its equilibrium position (in second). (Take `m=pi^2kg`,`K_2k=4K_1`,`K_1=17(N)/(m).`)

Explanation

(1) Net sunstituting in the tension in string connecting the block will provide acceleration to the block. We can write
`triangleT=ma` .(i)
When the block `m` is displaced by a distance `x` beyond equilibrium position, the addition stretch of springs `1` and `2` are `x_1` and `x_2` respectively, we can write
`x_1=(x-x_2)/(2)impliesx=2x_1+x_2` .(ii)
At `triangleT=k_2x_2` and `2triangleT=k_1x_1` then `2(k_2x_2)=k_1x_1`
`impliesx_1=(2k_2x_2)/(k_1)` .(iii)
from eqs. (ii) and (iii) `x=2[(2k_2x_2)/(k_1)]+x^2=((4k_2+k_1))/(k_1)x_2`
`impliesx_2=(k_1x)/((k_1+4k_2))`
and `veca=(-k_1k_2)/((k_1+4k_2)m).x`i `impliesveca=(-k_1k_2)/((k_1+4k_2)m).x`
`omega^2==(k_1k_2)/((k_1+4k_2)m)`
Hence `R=(80)/(23pi^2)m`
After substituting the values, we get `T=1s`