A circular spring of natural length `l_(0)` is cut and weided with two beads of masses `m_(1) and m_(2)` such that the ratio of the lengths of the springs between the beads is 4:1 if the stiffness of the spring is k. find the frequency of oscillation of the beads in a smooth horizontal rigid tube. Assume m_(1) = m and m_(2) = 3 m`.

Explanation

When `m_(1)` is displaced x, each spring will be deformedby same amoun. Hence, the springs are connected in parallel. The equivalent spring constant is
`k_(eq) = k_(1) + k_(2)`

If the spring is cut, the force constant of spring
`k prop (1)/(l) implies k_(1) l_(1) = k_(2) l_(2) = k l`
Substituting `l_(1) = l//5 and l_(2) = 4//5`, we have
`k_(1) = 5k and k_(2) = (5)/(4) k`
Then `k_(eq) = (25)/(4) k`
Now we have two partical of masses `m_(1) and m_(2)` and one spring of stiffness
`k_(eq) = (25)/(4) k`
The reduced mass is `mu = (m_(1) m_(2))/((m_(1) + m_(2))`
where `m_(1) = m and m_(2) = 3m`
This gives `mu = 3//4m`
Substituting `mu = 3//4m and k_(eq) = 25//4 k` in the formula
`omega = sqrt((k_(eq))/(mu)) implies omega = sqrt(((25)/(4)k)/((3)/(4)m)) = sqrt((25k)/(3m)) = 5 sqrt((k)/(3 m))`