A particle is performing $SHM$ according to the equation $x = (3\, cm)$ $\sin \,\left( {\frac{{2\pi t}}{{18}} + \frac{\pi }{6}} \right)$ where $t$ is in seconds. The distance travelled by the particle in $36\, s$ is ..... $cm$
A$24$
B$1.5$
C$25.5$
D
None of these
Medium
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A$24$
a $\omega t=\frac{2 \pi}{18} t \Rightarrow \omega=\frac{2 \pi}{18} \quad$ also $\quad \omega=\frac{2 \pi}{\mathrm{T}}$
$\because T=18 s$
So distance travelled in $36 \mathrm{sec}=8 \mathrm{A}$
$=8 \times 3=24 \mathrm{cm}$
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