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  4. The motion of a particle executing SHM is given by x=0.01sin 100 π (. t + 0.05 .) , where x is in metre and time t is in second. The time period is

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Q. The motion of a particle executing SHM is given by $x=0.01sin 100 \pi \left(\right. t + 0.05 \left.\right)$ , where $x$ is in metre and time $t$ is in second. The time period is

NTA AbhyasNTA Abhyas 2020Oscillations

Solution:

$x=0.01sin 100 \pi \left(t + 0.05\right)$
$=0.01sin \left(\right. 100 \pi t + 5 \pi \left.\right)$
$\therefore $ Angular frequency $\omega =100\pi =\frac{2 \pi }{T}$
or $T=\frac{2}{100}=0.02 \, $ s