A particle performs simple harmonic motion with a period of 2 second. The time taken by the particle to cover a displacement equal to half of its amplitude from the mean position is ${1 \over a}$s. The value of 'a' to the nearest integer is _________.
Answer (integer)
6
Solution
Time period (T) = 2 sec.<br><br>X = A sin ($\omega$t + $\phi$) ($\phi$ = 0 at M.P.)<br><br>$\Rightarrow$ ${A \over 2} = A\sin {{2\pi } \over T}t$<br><br>$\Rightarrow$ ${{2\pi } \over 2}t = {\pi \over 6}$<br><br>$\Rightarrow$ $t = {1 \over 6}$<br><br>$\therefore$ a = 6
About this question
Subject: Physics · Chapter: Oscillations · Topic: Simple Harmonic Motion
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