Question

A particle executing simple harmonic motion with an amplitude 5 cm and a time period 0.2 s. The velocity and acceleration of the particle when the displacement is 5 cm is

• A. 0.5$\pi$ m s–1, 0 m s–2
• B. 0.5 m s–1, –5$\pi$2 m s–2
• C. 0 m s–1, –5$\pi$2 m s–2
• D. 0.5$\pi$ m s–1, –0.5$\pi$2 m s–2

Answer 1

Answer: (C)

0 m s–1, –5$\pi$2 m s–2

Answer 2

Here, $A = 5cm = 0.05m,T = 0.2s$

$\therefore\omega = \frac{2\pi}{T} = \frac{2\pi}{0.2} = 10\pi rads^{- 1}$

Velocity and acceleration of the particle executing SHM at any displacement x is given by

Velocity $v = \omega\sqrt{A^{2} - x^{2}}$

And acceleration $a = - \omega^{2}x$

When x = 5 cm $= 0.05m$

$\therefore v = 10\pi\sqrt{(0.05)^{2} - (0.05)^{2}} = 0$

And $a = - (10\pi)^{2}(0.05) = - 5\pi^{2}ms^{- 2}$