I found the time period in both as $\pi /10$ and $3\pi /10$

If I assume one block at rest then the relative time period would be $\pi /5 $

After that I got confused .

Answer s given as $3\pi /10$

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The only physics here is finding the period of each oscillator, using $\omega=\sqrt{\frac{k}{m}}$. These are $\omega_1=20 rad/s$ and $\omega_2=\frac{20}{3}rad/s$. The rest is mathematics.

When in difficulty with calculations, you can always sketch or plot a graph (below). The displacements are $x_1=10\cos(20t)$ and $x_2=10\cos(\frac{20}{3}t)$.

My own initial confusion is understanding what is meant by being *again side by side*. It could mean that the blocks are :

1. passing and moving in the opposite direction; or

2. moving in the same direction , but not necessarily back in the start position; or

3. back at the start position.

These 3 options are illustrated in the graph above.

There is no way of knowing which time is intended. The times for each case are 1. $\frac34 \frac{\pi}{10}$ seconds, 2. $\frac32 \frac{\pi}{10}$ seconds, and 3. $3\frac{\pi}{10}$ seconds.

To get the same results by calculation we set $x_1=x_2$ :

$10\cos(\omega_1 t)=10\cos(\omega_2 t)$

$\omega_1 t = 2\pi n \pm \omega_2 t$

$(\omega_1 \pm \omega_2)t=2\pi n$

where $n$ is an integer, and I have used the fact that the cosine has reflection symmetry $(\cos(-x)=\cos(x))$ as well as periodic translation symmetry $(\cos(x)=\cos(x+2\pi n))$.

Taking the + sign we get for $n=1$ that

$\frac{80}{3}t=2\pi$

$t=\frac{3\pi}{40}$.

Taking the - sign we get

$\frac{40}{3} t=2\pi$

$t=\frac{3 \pi}{20}$.

Taking the - sign with $n=2$ we get

$\frac{40}{3}t=4\pi$

$t=\frac{3 \pi }{10}$.

Alternatively, starting from your calculation of the periods $T_1=\frac{\pi}{10}$ and $T_2=\frac{3\pi}{10}$ :

The blocks will return to the starting position at the same time when each block has made a whole number of cycles. The time after which the blocks reach this position is

$t=mT_1=nT_2$

$\frac{m}{n}=\frac{T_2}{T_1}=3$.

The smallest possible integer values are $m=3, n=1$. Then $t=nT_2=\frac{3\pi}{10}$.

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