Question

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

Answer 1

Keff $= K + K$ as both springs are in use in parallel
$=2 k $
$=2 \times 2=4 N / m $
$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
$N =\frac{\text { time }}{ T }=\frac{14 \pi}{7 \pi / 10}=20$