As shown in the figure, S₁ and S₂ are identical springs with spring constant K each. The oscillation frequency of the mass ' m ' is ' f '. If the spring S₂ is removed, the oscillation frequency will become

Step-by-step solution

Effective Spring Constants and Frequency Relationships For two identical springs with spring constant K attached in parallel to mass m , the effective spring constant is the sum: K_ eff , initial = K + K = 2K The oscillation frequency is given by: f = 1 2 2K m When one spring is removed, the system has a single spring with constant K : K_ eff , new = K The new frequency is: f' = 1 2 K m Comparing the two expressions: f = 2 f' The new frequency is therefore: f' = f 2 The correct answer is option C.