Question 156 of 194

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

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Answer: 20

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