1. Tardigrade
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  4. A particle executes SHM of type x=A sin ω t. It takes time t1 from x=0 to x=(A/2) and t2 from x=(A/2) to x=A . The ratio t1: t2 will be

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Q. A particle executes SHM of type $x=A \sin \omega t$. It takes time $t_{1}$ from $x=0$ to $x=\frac{A}{2}$ and $t_{2}$ from $x=\frac{A}{2}$ to $x=A .$ The ratio $t_{1}: t_{2}$ will be

Oscillations

Solution:

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$t_{1}+t_{2}=\frac{T}{4}$
or $ t_{2}=\frac{T}{4}-t_{1}$
At time, $t=t_{1}, x=\frac{A}{2}$
$\therefore \,\,\,\,\frac{A}{2}=A \sin \omega t_{1}$ or $\omega t_{1}=\frac{\pi}{6} \quad$ or $t_{1}=\frac{\pi}{6 \omega}$
$\therefore \,\,\,\,t_{2}=\frac{T}{4}-\frac{\pi}{6 \omega}=\frac{2 \pi}{4 \omega}-\frac{\pi}{6 \omega}=\frac{2 \pi}{6 \omega} \,\,\,\,\,\left(\because T=\frac{2 \pi}{\omega}\right)$
$\therefore t_{1}: t_{2}=1: 2$