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(C) Given, the displacement equation of the first particle is $y_1=30 \sin \left(2 \pi t+\frac{\pi}{3}\right)$.Comparing this with the standard $SHM$

Vedclass · Accepted Answer

$3: 2$

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(C) Given, the displacement equation of the first particle is $y_1=30 \sin \left(2 \pi t+\frac{\pi}{3}ight)$.Comparing this with the standard $SHM$ equation $y=A \sin(\omega t + \phi)$, the amplitude is $A_1=30$.For the second particle, the displacement equation is $y_2=10(\sin 2 \pi t+\sqrt{3} \cos 2 \pi t)$.To find the amplitude, we multiply and divide by $2$ inside the bracket: $y_2 = 10 	imes 2 \left[ \frac{1}{2} \sin 2 \pi t + \frac{\sqrt{3}}{2} \cos 2 \pi t ight]$.Using the trigonometric identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$, where $\cos \frac{\pi}{3} = \frac{1}{2}$ and $\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$, we get $y_2 = 20 \sin \left(2 \pi t + \frac{\pi}{3} ight)$.Thus, the amplitude of the second particle is $A_2=20$.The ratio of the amplitudes is $\frac{A_1}{A_2} = \frac{30}{20} = \frac{3}{2}$.Therefore, the ratio $A_1: A_2$ is $3: 2$.

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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