Two equal masses are connected by a spring satisfying Hook's law and are placed on a frictionless table. The spring is elongated a little and allowed to go. Let the angular frequency of oscillations be `omega`. Now one of the masses is stopped. The square of the new angular frequency is `:`

Options

A. `omega^(2)`
B. `(omega^(2))/(2)`
C. `(omega^(2))/(3)`
D. `2 omega^(2)`

Answer

B

Explanation

`sqrt((k)/(mu))=omega rArr mu=(m_(1)m_(2))/(m_(1)+m_(2))`
`omega =sqrt((2k)/(m))rArr omega_(2)sqrt((k)/(m))`
`omega_(2)^(2) =(k)/(m)=(omega^(2))/(2)`