A simple harmonic motion oscillator has two springs of spring constant k and other two of spring constant 2k, arranged as shown in the figure. Calculate the time period of oscillator.

Explanation

The parallel combination of upper two springs (springs constant k) is connected in series with parallel combination of lower two springs (spring constant 2k). So, the equivalent spring constant of the oscillator is
`(1)/(k_(eq)) = ((1)/(k + k)) + ((1)/(2k + 2k)) = (1)/(2k) + (1)/(4k) = (3)/(4k)`
`rArr k_(eq) = (4k)/(3)`
`therefore` Time period of oscillator is
`T = 2pi sqrt((m)/(k_(eq))) = 2pisqrt((3m)/(4k)) = pisqrt((3m)/(k))`