Question

All the surfaces are smooth and springs are ideal. If a block of mass m is given the velocity v0 to the left direction, then the time period of block as shown in the figures is​

Answer 1

The time period of the block is

$\boxed{\frac{2(l_1+l_2)}{v_0}+\frac{3\pi}{2}\sqrt{\frac{m}{K}}}$

Explanation:

The time period will be equal to the time taken by the block to come back to its original position

The block will compress spring with spring constant K and then spring will come back to its original position

This time will be equal to half of the time period of spring constant K

Similar thing can be said for spring with spring constant 4K

Therefore, time period of the block

$T=\frac{1}{2}(2\pi\sqrt{\frac{m}{K}}+2\pi\sqrt{\frac{m}{4K}})+2\frac{l_1}{v_0}+2\frac{l_2}{v_0}$

$\implies T=\pi(\sqrt{\frac{m}{K}}+\frac{1}{2}\sqrt{\frac{m}{K}})+\frac{2(l_1+l_2)}{v_0}$

$\implies T=\frac{3\pi}{2}\sqrt{\frac{m}{K}}+\frac{2(l_1+l_2)}{v_0}$

This is the time period of the block.

Hope this answer is helpful.

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