A load of mass `M` is attached to the bottom of a spring of mass `'M //3'` and spring constant 'K'. If the system is set into oscillation, the time period of oscillation is

Options

A. `(2pi)/(3) sqrt((10 M)/(3K))`
B. `4 pi sqrt((M)/(3K))`
C. `4pi sqrt((M)/(K))`
D. `2 pi sqrt((M)/(3K))`

Answer

A

Explanation

Effective mass of a spring `(1)/(3)` actual mass
`= (1)/(3) ((M)/(3)) = (M)/(9)`
Total mass `= M + (M)/(9) = (10 M)/(9)`
`T = 2pi sqrt(((10 M)/(9))/(k)) = (2pi)/(3) sqrt((10 M)/(k))`