Two springs are jonied and connected to a mass m as shown in figure. If the force constants of the two springs are `k_(1) and k_(2)`, shown that frequency of oscillation of mass m is

Options

A. `(1)/(2pi)sqrt((k_(1)k_(2))/((k_(1)+k_(2))m))`
B. `2pisqrt(((k_(1)+k_(2))m)/(k_(1)k_(2)))`
C. `2pisqrt((k_(1)k_(2))/(m))`
D. `(1)/(2pi)sqrt((m)/(k_(1)+k_(2)))`

Answer

C

Explanation

Let the mass m be displaced by a small distance x towards right. Then `x=-(F)/(k)`
If `x_(1)andx_(2)` are the extensions in the two springs, then
`x_(1)=-(F)/(k_(1))andx_(2)=-(F)/(k_(2))` But `x=x_(1)+x_(2)`
`therefore -(F)/(k)=-(F)/(k_(1))-(F)/(k_(2))or(1)/(k)=(1)/(k_(1))+(1)/(k_(2))`
or `(1)/(k)=(k_(2)+k_(1))/(k_(1)k_(2))ork=(k_(1)k_(2))/(k_(1)+k_(2))`
`therefore upsilon=(1)/(2pi)sqrt((k_(1)k_(2))/((k_(1)+k_(2))m))`