The equation of a particle executing simple harmonic motion is given by $x = \sin \pi \left( {t + {1 \over 3}} \right)m$. At t = 1s, the speed of particle will be
(Given : $\pi$ = 3.14)
Solution
<p>$x = \sin \left( {\pi t + {\pi \over 3}} \right)m$</p>
<p>$\Rightarrow {{dx} \over {dt}} = \pi \cos \left( {\pi t + {\pi \over 3}} \right)$</p>
<p>$= \pi \cos \left( {\pi + {\pi \over 3}} \right)$ at $t = 1\,s$</p>
<p>$= - {\pi \over 2}$ m/s</p>
<p>or $\left| {{{dx} \over {dt}}} \right| = 157$ cm/s</p>
About this question
Subject: Physics · Chapter: Oscillations · Topic: Simple Harmonic Motion
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