Question

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 $\left(K=2N{m}^{-1}\right)$. 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

Answer: 20

Keff $=K+K$ as both springs are in use in parallel
$=2k$
$=2\times 2=4N/m$
$m=490gm$
$=0.49kg$
$T=2\pi \sqrt{\frac{m}{Keff}}=2\pi \sqrt{\frac{0.49kg}{4}}$
$=2\pi \sqrt{\frac{49}{400}}=2\pi \frac{7}{20}=\frac{7\pi }{10}$
No. of oscillation in the $14\pi$ is
$N=\frac{\text{ time }}{T}=\frac{14\pi }{7\pi /10}=20$