(A) The effective spring constant $K_{\text{eff}} = K + K$ because both springs are in parallel configuration.$K_{\text{eff}} = 2K = 2 \times 2 = 4 \,

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(A) The effective spring constant $K_{\text{eff}} = K + K$ because both springs are in parallel configuration.$K_{\text{eff}} = 2K = 2 \times 2 = 4 \, N \, m^{-1}$.The mass of the block is $M = 490 \, g = 0.49 \, kg$.The time period $T$ of the oscillation is given by $T = 2 \pi \sqrt{\frac{M}{K_{\text{eff}}}}$.Substituting the values: $T = 2 \pi \sqrt{\frac{0.49}{4}} = 2 \pi \sqrt{\frac{49}{400}} = 2 \pi \left( \frac{7}{20} \right) = \frac{7 \pi}{10} \, s$.The number of complete oscillations $N$ in time $t = 14 \pi \, s$ is $N = \frac{t}{T}$.$N = \frac{14 \pi}{7 \pi / 10} = 14 \pi \times \frac{10}{7 \pi} = 20$.

Class 11 Physics Oscillations SHM of Spring Mass System

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In the figure given below,a block of mass $M = 490 \, g$ placed on a frictionless table is connected with two springs having the 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 $14 \pi$ seconds will be $.........$

JEE MAIN 2023 All JEE MAIN

A
$20$
B
$21$
C
$19$
D
$26$

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