JEE MAIN - Physics (2023 - 6th April Morning Shift - No. 13)
A mass $$m$$ is attached to two strings as shown in figure. The spring constants of two springs are $$\mathrm{K}_{1}$$ and $$\mathrm{K}_{2}$$. For the frictionless surface, the time period of oscillation of mass $$m$$ is :
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$$2\pi \sqrt {{m \over {{K_1} + {K_2}}}} $$
$$2\pi \sqrt {{m \over {{K_1} - {K_2}}}} $$
$${1 \over {2\pi }}\sqrt {{{{K_1} + {K_2}} \over m}} $$
$${1 \over {2\pi }}\sqrt {{{{K_1} - {K_2}} \over m}} $$
Explanation
Since, both spring are in parallel
Combination, therefore, $K_{\text {eq }}=K_1+K_2$
Time period of oscillation, $T=2 \pi \sqrt{\frac{m}{K_{\text {eq }}}} $
$\Rightarrow T=2 \pi \sqrt{\frac{m}{K_1+K_2}}$
Option (a) is correct.
Combination, therefore, $K_{\text {eq }}=K_1+K_2$
Time period of oscillation, $T=2 \pi \sqrt{\frac{m}{K_{\text {eq }}}} $
$\Rightarrow T=2 \pi \sqrt{\frac{m}{K_1+K_2}}$
Option (a) is correct.
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