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QJEEM23Ph10dc: JEE Main 2023 Physics - Class 11 JEEM23Ph10dc Solution: In the figure given below, a block of mass M = 490   g…

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 $(K = 2 N m^{- 1})$ . If the block is horizontally displaced through ' $X$ ' $m$ then the number of complete oscillations it will make in $14 π$ seconds will be _____________.

JEE Main 2023 (Online) 31st January Morning Shift Physics - Simple Harmonic Motion Question 42 English

Step-by-Step Verified Solution

\mathrm{K_eff}=\mathrm{K}+\mathrm{K}

=2 \mathrm{k}

(As both springs are in use in parallel)

=2 \times 2=4 \mathrm{~N} / \mathrm{m}

m = 490 gm

\begin{aligned} = 0.49 kg \end{aligned}

\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{Keff}}}=2 \pi \sqrt{\frac{0.49 \mathrm{~kg}}{4}}

= 2 π \sqrt{\frac{49}{400}} = 2 π \frac{7}{20} = \frac{7 π}{10}

No. of oscillation in the

14 \pi

is

N = \frac{time}{T} = \frac{14 π}{7 π / 10} = 20

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