Two particles A and B of equal masses are suspended from two massless springs of spring constants `k_(1)` and `k_(2)`, respectively, If the maximum velocities, during oscillation, are equal, the ratio of amplitude of A and B is

Options

A. `sqrt((k_(1))/(k_(2)))`
B. `(k_(2))/(k_(1))`
C. `sqrt((k_(2))/(k_(1)))`
D. `(k_(1))/(k_(2))`

Answer

C

Explanation

Maximum velocity during SHM = `A omega`
But `k=m omega^(2)`
`:." "omega=sqrt((k)/(m))" ":." Maximum velocity "A sqrt((k)/(m))`
Here the maximum velocity is same and m is also same
`:." "A_(1)sqrt(k_(1))=A_(2)sqrt(k_(2))" ":.(A_(1))/(A_(2))=sqrt((k_(2))/(k_(1)))`