A particle is executing SHM along a straight line. Its velocities at distances $ \mathrm{x}_{1} $ and $ \mathrm{x}_{2} $ from the mean position are $ \mathrm{V}_{1} $ and $ \mathrm{V}_{2} $ respectively. Its time period is:

Question

A particle is executing SHM along a straight line. Its velocities at distances $ \mathrm{x}_{1} $ and $ \mathrm{x}_{2} $ from the mean position are $ \mathrm{V}_{1} $ and $ \mathrm{V}_{2} $ respectively. Its time period is:, $ 2 \pi \sqrt{\frac{x_{1}^{2}+x_{2}^{2}}{V_{1}^{2}+V_{2}^{2}}} $, $ 2 \pi \sqrt{\frac{x_{2}^{2}-x_{1}^{2}}{v_{1}^{2}-v_{2}^{2}}} $, $ 2 \pi \sqrt{\frac{\mathrm{v}_{1}^{2}+\mathrm{V}_{2}^{2}}{\mathrm{x}_{1}^{2}+\mathrm{x}_{2}^{2}}} $, $ 2 \pi \sqrt{\frac{\mathrm{v}_{1}^{2}-\mathrm{V}_{2}^{2}}{\mathrm{x}_{1}^{2}-\mathrm{x}_{2}^{2}}} $

MARKS App · Accepted Answer

For particle undergoing SHM, V = ω A 2 - x 2 ⇒ V 2 = ω 2 A 2 - x 2 V 1 = ω A 2 - x 1 2 ⇒ V 1 2 = ω 2 A 2 - x 1 2 ...(i) V 2 = ω A 2 - x 2 2 ⇒ V 2 2 = ω 2 A 2 - x 2 2 ...(ii) V 1 2 - V 2 2 = ω 2 x 2 2 - x 1 2 ω = V 1 2 - V 2 2 x 2 2 - x 1 2 T = 2 π x 2 2 - x 1 2 V 1 2 - V 2 2

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