In the figure given below. a block of mass M = 490g placed on a...

Both the springs are connected in parallel, therefore

${K}_{eq} = K + K = 2K$

Now, time period of spring-block system is given by,

$T = 2\pi\sqrt{\dfrac{m}{{K}_{eq}}} = 2\pi\sqrt{\dfrac{m}{2K}}$

Given here, $m = 490gm$$= 0 . 49kg$ and $K = 2N{m}^{- 1}$

So, $T = 2\pi\sqrt{\dfrac{0 . 49}{2 \times 2}}$

$= 2\pi\sqrt{\dfrac{49}{400}} = 2\pi\dfrac{7}{20} = \dfrac{7\pi}{10}$

Now, number of oscillation in $14\pi$ is

$N = \dfrac{time}{T} = \dfrac{14\pi}{\dfrac{7\pi}{10}} = 20$