Two springs, of spring constants `k_(1)` and `K_(2)`, have equal highest velocities, when executing SHM. Then, the ratio of their amplitudes (given their masses are in the ratio `1:2`) will be

Options

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

Answer

A

Explanation

The angular frequency of spring is given by
`omega = sqrt((k)/(m)) prop sqrt(k)`
For equal maximum velocities, we have
`A_(1)omega_(1) = A_(2)omega_(2)`
or `(A_(1))/(A_(2))=(omega_(2))/(omega_(1))=sqrt((k_(2))/(k_(1)))=((k_(2))/(k_(1)))^((1)/(2))" "(because m = m_(1) = m_(2))`