Q142 / 160 · TS EAMCET 2019 (04 May Shift 2) · Physics

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

  • A
    $2: 1$
  • B
    $1: 1$
  • Correct answer
    C
    $3: 2$
  • D
    $1: \sqrt{3}$

Solution

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, $$ \begin{aligned} y_2 & =10(\sin 2 \pi t+\sqrt{3} \cos 2 \pi t) \\ & =10 \times 2\left[\frac{1}{2} \sin 2 \pi t+\frac{\sqrt{3}}{2} \cos 2 \pi t\right] \\ & =20\left[\cos \frac{\pi}{3} \sin 2 \pi t+\sin \frac{\pi}{3} \cos 2 \pi t\right] \\ \Rightarrow \quad y_2 & =20 \sin \left(2 \pi t+\frac{\pi}{3}\right) \end{aligned} $$

Hence, the correct answer is option: 'C'.

Relevant Questions