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Two springs having stiffness k1 and k2, respectively, are used in a vertical position to

support a single object of mass m. Show that the angular frequency of oscillation is
[(k1 + k2) /m]¹/² if the springs are tied in parallel, and [k1k2 / (k1 + k2)m]¹/² if the springs
are tied in series.

1 Answer

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Final answer:

In parallel, the combined stiffness is the sum of individual stiffnesses, leading to angular frequency ω = √[(k1 + k2)/m]. In series, the effective stiffness is computed as the harmonic mean of the individual stiffnesses, leading to angular frequency ω = √[k1k2 / (k1 + k2)m].

Step-by-step explanation:

The question deals with two scenarios of oscillation involving springs: one where the springs are connected in parallel, and the other where they are connected in series. In the parallel scenario, the force constants (spring stiffnesses) of the springs are simply added because they both exert force on the mass simultaneously. The resultant stiffness, kp = k1 + k2, is used to calculate the angular frequency of oscillation using the formula ω = √(k/m), leading to ω = √[(k1 + k2)/m]. In the series scenario, the effective stiffness of the springs is given by the reciprocal of the sum of the reciprocals of their stiffnesses, such that 1/ks = 1/k1 + 1/k2. This new composite spring constant leads to an angular frequency of ω = √[k1k2 / (k1 + k2)m].

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