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 $\left( K =2 N m ^{-1}\right)$. If the block is horizontally displaced through ' $X$ 'm then the number of complete oscillations it will make in $14 \pi$ seconds will be $.........$
JEE MAIN 2023, Medium
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$Keff = K + K$ as both springs are in use in parallel
$=2\,k$
$=2 \times 2=4\,N / m \quad m =490\,gm$
$=0.49\,kg$
$T =2 \pi \sqrt{\frac{ m }{ Keff }}=2 \pi \sqrt{\frac{0.49\,kg }{4}}$
$=2 \pi \sqrt{\frac{49}{400}}=2 \pi \frac{7}{20}=\frac{7 \pi}{10}$
No. of oscillation in the $14 \pi$ is
$N =\frac{\text { time }}{ T }=\frac{14 \pi}{7 \pi / 10}=20$
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