A particle of mass 0.1 kg is held between two rigid supports by two springs of force constant `8 N m^(-1) and 2 N m^(-1)`. If the particle is displaced along the direction of length of the springs, its frequency of vibration is

Options

A. `(5)/(pi)Hz`
B. `(8)/(pi)Hz`
C. `(2)/(pi)Hz`
D. `(1)/(pi)Hz`

Answer

A

Explanation

Here, `m=0.1kg,k_(1)=8N//m,k_(2)=2N//m`

The spring are connected in parallel, the equivalent spring constant is
`k_(eq)=k_(1)+k_(2)=8N//m+2N//m=10N//m`
The frequency of oscillation is
`upsilon=(1)/(2pi)sqrt((k_(eq))/(m))=(1)/(2pi)sqrt(((10N//m))/((0.1kg)))=(10)/(2pi)Hz=(5)/(pi)Hz`