Four massless springs whose force constants are 2k,2k,k and 2k respectively are attached to a mass M kept on a frictionless plane (as shown in figure). If the mass M is displaced slightly in the horizontal direction, then the frequency of oscillation of the system is,
Springs on the left of the block are in series, hence their equivalent spring constant is
K1=(2K)(2K)2K+2K=K
Spring on the right of the block are in parallel, hence their equivalent spring constant is
K2=K+2K=3K
Again both K1 and K2 are in parallel
∴Keq=K1+K2=K+3K=4K
Hence, frequency is f=12π√KeqM=12π√4KM