In the figure given below. a block of mass $M = 490 g$ placed on a frictionless table is connected with two springs having same spring constant ( $K = 2 N m^{- 1}$ ). If the block is horizontally displaced through $X$ $m$ then the number of complete oscillations it will make in $14 π$ seconds will be ______.

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TestHub · Accepted Answer

Both the springs are connected in parallel, therefore

$K_{e q} = K + K = 2 K$

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

$T = 2 π \sqrt{\frac{m}{K_{e q}}} = 2 π \sqrt{\frac{m}{2 K}}$

Given here, $m = 490 gm$ $= 0.49 kg$ and $K = 2 N m^{- 1}$

So, $T = 2 π \sqrt{\frac{0.49}{2 \times 2}}$

$= 2 π \sqrt{\frac{49}{400}} = 2 π \frac{7}{20} = \frac{7 π}{10}$

Now, number of oscillation in $14 π$ is

$N = \frac{time}{T} = \frac{14 π}{\frac{7 π}{10}} = 20$

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