Two springs of force constants `k_(1)` and `k_(2)`, are connected to a mass m as shown. The frequency of oscillation of the mass is `f`. If both `k_(1)` and `k_(2)` are made four times their original values, the frequency of oscillation becomes

Options

A. `2upsilon`
B. `upsilon//2`
C. `upsilon//4`
D. `4upsilon`

Answer

A

Explanation

In the given figure two springs are connected in parallel. Therefore the effective spring constant is given by

`k_(eff)=k_(1)+k_(2)`
Frequency of oscillation,
`upsilon=(1)/(2pi)sqrt((k_(eff))/(m))=(1)/(2pi)sqrt((k_(1)+k_(2))/(m))" "...(i)`
As `k_(1)andk_(2)` are increased four times
New frequency,
`upsilon.=(1)/(2pi)sqrt((4(k_(1)+k_(2)))/(m))=2upsilon` (using (i))