JEE MAIN - Physics (2022 - 25th July Morning Shift - No. 9)

JEE Main 2022 (Online) 25th July Morning Shift Physics - Simple Harmonic Motion Question 53 English

In figure $$(\mathrm{A})$$, mass '$$2 \mathrm{~m}^{\text {' }}$$ is fixed on mass '$$\mathrm{m}$$ ' which is attached to two springs of spring constant $$\mathrm{k}$$. In figure (B), mass '$$\mathrm{m}$$' is attached to two springs of spring constant '$$\mathrm{k}$$' and '$$2 \mathrm{k}^{\prime}$$. If mass '$$\mathrm{m}$$' in (A) and in (B) are displaced by distance '$$x^{\prime}$$ horizontally and then released, then time period $$\mathrm{T}_{1}$$ and $$\mathrm{T}_{2}$$ corresponding to $$(\mathrm{A})$$ and (B) respectively follow the relation.

$$ \frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\frac{3}{\sqrt{2}} $$

$$ \frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\sqrt{\frac{3}{2}} $$

$$ \frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\sqrt{\frac{2}{3}} $$

$$ \frac{T_{1}}{T_{2}}=\frac{\sqrt{2}}{3} $$

Explanation

Both the springs are in parallel combination in both the diagrams so

$${T_1} = 2\pi \sqrt {{{3m} \over {2k}}} $$

and $${T_2} = 2\pi \sqrt {{m \over {3k}}} $$

So, $${{{T_1}} \over {{T_2}}} = {3 \over {\sqrt 2 }}$$

JEE MAIN - Physics (2022 - 25th July Morning Shift - No. 9)

Explanation

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