The function of time representing a simple harmonic motion with a period of ${\pi \over \omega }$ is :
Solution
General equation of SHM<br><br>x = A sin($\omega$'t $\pm$ $\phi$)<br><br>We know, $\omega$ = ${{2\pi } \over T}$<br><br>Given, $T = {\pi \over \omega }$<br><br>$\therefore$ $\omega$' = ${{2\pi } \over {{\pi \over \omega }}}$ = 2$\omega$<br><br>$\therefore$ Equation becomes,<br><br>x = a sin(2$\omega$t $\pm$ $\phi$)<br><br>Here coefficient of t is 2$\omega$.<br><br>you can see only option (D) has coefficient 2$\omega$.
About this question
Subject: Physics · Chapter: Oscillations · Topic: Simple Harmonic Motion
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