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(B) For $S.H.M.$,the force on the particle is given by $F = -kx$.Since $F = ma$,we have $ma = -kx$,which implies $a = -(k/m)x$.Comparing this with the

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All motions having the same time period are $S.H.M.$

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(B) For $S.H.M.$,the force on the particle is given by $F = -kx$.Since $F = ma$,we have $ma = -kx$,which implies $a = -(k/m)x$.Comparing this with the standard $S.H.M.$ equation $a = -\omega^2 x$,we get $\omega^2 = k/m$,which is a constant.Since the angular frequency $\omega$ is constant,the time period $T = 2\pi/\omega$ is also constant. Thus,statement $A$ is correct.Not all periodic motions are $S.H.M.$ For example,uniform circular motion has a fixed time period but is not $S.H.M.$ because the force is not proportional to displacement. Thus,statement $B$ is incorrect.The total energy of $S.H.M.$ is given by $E = (1/2)kA^2$,which shows that energy is directly proportional to the square of the amplitude $(A^2)$. Thus,statement $C$ is correct.The displacement equation is $x = A \sin(\omega t + \phi)$,where $\phi$ is the phase constant. The value of $\phi$ is determined by the initial position and velocity of the particle at $t = 0$. Thus,statement $D$ is correct.Therefore,the wrong statement is $B$.

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