Q. 14.24 (c) A body describes simple harmonic motion with an amplitude of 5\; cm and a period of 0.2\; s. Find the acceleration and velocity of the body when the displacement is 

(c)  0 \; cm

Answers (1)
S Sayak

A = 5 cm = 0.05 m

T = 0.2 s

\\\omega =\frac{2\pi }{T}\\ \omega =\frac{2\pi }{0.2}\\ \omega =10\pi\ rad\ s^{-1}

At displacement x acceleration is a=-\omega ^{2}x

At displacement x velocity is v=\omega \sqrt{A^{2}-x^{2}}

(a)At displacement 0 cm

\\v=10\pi \sqrt{(0.05)^{2}-(0)^{2}}\\ v=10\pi \times 0.05\\v=1.57ms^{-1}

\\a=-(10\pi )^{2}\times0\\a=0

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