As the cart is moving with a constant acceleration
#veca# in horizontal direction, here the block of mass M is assumed to experience continually a pseudo force, also called fictitious force, of magnitude #Mveca# in the direction opposite to the direction of acceleration of the cart. The reaction force thus produced on springs causes expansion of one spring and compression of the other. As such restoring forces will be operative on the springs. All these forces will provide the mass M an equilibrium position in accelerated frame of reference. This equilibrium position of the mass differs from the equilibrium position of the mass, if the cart has got no acceleration.
Let this equilibrium position of the center of mass of M be O. Now if the center of mass of M be pushed rightward by a small distance x along +ve direction of X-axis from this equilibrium position O,then the left spring will be expanded by a length x and right spring will be compressed by the same length x.Restoring force on both the spring will act along negative direction of X-axis.
So considering the net restoring force on the mass M due to the displacement of its center of mass by an amount x rightward we can write
#M(d^2x)/(dt^2)=-k_1x-k_2x#
#=>M(d^2x)/(dt^2)=-(k_1+k_2)x#
#=>(d^2x)/(dt^2)=-(k_1+k_2)/Mx#
This equation shows that the mass M will execute a SHM where the angular frequency #omega# of the paricle moving in reference circle associated with the SHM will be
#omega=sqrt((k_1+k_2)/M)#
If the frequency of SHM be #n# then we can write from above equation
#omega=2pin=sqrt((k_1+k_2)/M)#
So frequency of SHM
#n=1/(2pi)sqrt((k_2+k_2)/M)#
