Q. A particle executes simple harmonic motion with a time period of 16 s. At time $t = 2$ s, the particle crosses the mean position while at $t = 4$ s, velocity is 4 $ms^{- 1}$. The amplitude of motion in metre is

A) $\sqrt{2}\pi$
B) $16\sqrt{2}\pi$
C) $24\sqrt{2}\pi$
D) $\frac{32\sqrt{2}}{\pi}$
Explanation

(d) For simple harmonic motion, $y = a{sin\ \ }{\omega t}$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ y = a{sin\ \ }{\left( \frac{2\pi}{T} \right)t}$ (at $t$=2 s)
$y_{1} = a{sin\ \ }\left\lbrack \left( \frac{2\pi}{16} \right) \times 2 \right\rbrack$
= $a\sin{\left( \frac{\pi}{4} \right) = \frac{a}{\sqrt{2}}}$ …(i)
At $t$=4 s or after 2 s from mean position.
$y_{1} = \frac{a}{\sqrt{2}}$ , velocity=4 $ms^{- 1}$
$\therefore$ Velocity = $\omega\sqrt{a^{2} - y_{1}^{2}}$
or $4 = \left( \frac{2\pi}{16} \right)\sqrt{a^{2} - \frac{a^{2}}{2}}$ [From Eq. (i)]
or 4 = $\frac{\pi}{8} \times \frac{a}{\sqrt{2}}$
or $a = \frac{32\sqrt{2}}{\pi}$ m

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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