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  • 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 (b) 3 cm

Q. 14.24 (b) 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

(b)  3\; cm

Answers (1)

best_answer

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 3 cm

\\v=10\pi \sqrt{(0.05)^{2}-(0.03)^{2}}\\ v=10\pi \sqrt{0.0016}\\v=10\pi \times 0.04\\v=1.257ms^{-1}

\\a=-(10\pi )^{2}\times 0.03\\a=-29.61ms^{-2}

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Sayak

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