Question

Two springs with constants k and 2k are connected in series and a mass m is hanged. The time period of oscillation is

Hint:

For springs in series, the equivalent spring constant is always less than the smallest individual spring constant. $k_{eq} = \frac{\text{product}}{\text{sum}} = \frac{k \cdot 2k}{k + 2k} = \frac{2k^2}{3k} = \frac{2k}{3}$.

Answer 1