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Two particles executing simple harmonic motion as described by $y_1=30 \sin \left(2 \pi t+\frac{\pi}{3}\right)$ and $y_2=10(\sin 2 \pi t+\sqrt{3} \cos 2 \pi t)$ have amplitudes $A_1$ and $A_2$ respectively. The ratio $A_1: A_2$ is
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Verified Answer
The correct answer is:
$3: 2$
Given,
displacement equation of the first particle,
$$
y_1=30 \sin \left(2 \pi t+\frac{\pi}{3}\right)
$$
Hence, its amplitude, $A_1=30 \mathrm{~m}$
displacement equation of the second particle,
$$
\begin{aligned}
y_2 & =10(\sin 2 \pi t+\sqrt{3} \cos 2 \pi t) \\
& =10 \times 2\left[\frac{1}{2} \sin 2 \pi t+\frac{\sqrt{3}}{2} \cos 2 \pi t\right] \\
& =20\left[\cos \frac{\pi}{3} \sin 2 \pi t+\sin \frac{\pi}{3} \cos 2 \pi t\right] \\
\Rightarrow \quad y_2 & =20 \sin \left(2 \pi t+\frac{\pi}{3}\right)
\end{aligned}
$$
displacement equation of the first particle,
$$
y_1=30 \sin \left(2 \pi t+\frac{\pi}{3}\right)
$$
Hence, its amplitude, $A_1=30 \mathrm{~m}$
displacement equation of the second particle,
$$
\begin{aligned}
y_2 & =10(\sin 2 \pi t+\sqrt{3} \cos 2 \pi t) \\
& =10 \times 2\left[\frac{1}{2} \sin 2 \pi t+\frac{\sqrt{3}}{2} \cos 2 \pi t\right] \\
& =20\left[\cos \frac{\pi}{3} \sin 2 \pi t+\sin \frac{\pi}{3} \cos 2 \pi t\right] \\
\Rightarrow \quad y_2 & =20 \sin \left(2 \pi t+\frac{\pi}{3}\right)
\end{aligned}
$$
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