Oscillations concept about Distance My question is regarding the distance covered by oscillator during 1 cycle of its SHM.According to my understanding it should be 4×a (where a=amplitude).Is this correct?
 A: I'll assume that you're talking about the simple case of the one-dimensional simple harmonic oscillator: $x(t) = A\cos(\omega t + \phi)$. For simplicity we'll assume $\phi = 0$**. Let $T$ be the period of oscillation. From Calculus we know that $ds^2 = dx^2 + dy^2 = dx^2 \implies ds = |dx|$ so
$$S = \int_T ds = \int_T |dx|\,.$$
We also know that $dx/dt = -A\omega\sin(\omega t)$ so
\begin{align*}
\int_T|dx| = \int_{0}^{2\pi/\omega}A\omega|\sin(\omega t)|dt &= A\omega\left[\int_0^{\pi/\omega}\sin(\omega t)dt +\int_{\pi/\omega}^{2\pi/\omega}-\sin(\omega t)dt\right]\\
&= A\omega\left[\frac{1}{\omega}(-(-1)-(-1)) + \frac{1}{\omega}(1-(-1))\right] \\
&= 4A\,.
\end{align*}
**This integral is different (not so sure any more, see other answer) if you don't assume $\phi = 0$. In order to get the same results, you would have to change the limits of integration. If you don't change the limits, you get a different answer depending on the phase shift (which is an interesting property of SHOs!). Hint: if you keep the limits the same, you should get an answer of the form $$ S = A[2\cos\phi - 2\cos(\pi + \phi)]\,.$$
A: Let the unit vector in the positive x-direction be $\hat i$
If the displacement of a particle undergoing simple harmonic motion of amplitude $A$ at a time is  $\vec x= A \sin \omega t \;\hat i$ then the total displacement of the particle over one period of the oscillation from time $t=0$ is $(+A \;\hat i)+  (-2A \;\hat i)+(+A \;\hat i) =0\hat i $.
The total distance travelled by the particle during that one period is $A+2A+A=4A$.  
Now suppose that the displacement is given by $\vec x= A \sin (\omega t+\phi) \;\hat i$, then the total displacement of the particle over one period of the oscillation from time $t=0$ is $([+A -A\sin \phi] \;\hat i)+ (-2A \;\hat i)+([+A+A\sin\phi]  \;\hat i)=0\;\hat i$.
The total distance travelled by the particle during that one period is $[A-A\sin \phi]+2A+[A+A\sin \phi]=4A$.  
So it does not matter which time interval of one period is considered, the displacement of the particle is always zero (the particle comes back to the same position as it started travelling in the same direction as when it started and the distance travelled by the particle is four times the amplitude.
