1.

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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