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A particle performing linear SHM has a period of 6.28 seconds and path length of 20 cm. What is the velocity when its displacement is 6 cm from the mean position? |
Answer» Data : T = 6.28 s, 2A = 20 cm ∴ A = 10 cm, x = 6 cm v = \(\omega\) \(\sqrt{A^2-x^2}\) = \(\frac{2\pi}T\)\(\sqrt{A^2-x^2}\) The velocity of the particle at x = 6 cm, \(\therefore\)v = \(\frac{2\times3.14}{6.28}\)\(\sqrt{(10)^2-(6)^2}\) = \(\sqrt{100-36}\) = \(\sqrt{64}\) = ± 8 cm/s |
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