The displacements of two particles executing simple harmonic motion are represented as $\mathrm{y}_{1}=2…

The displacements of two particles executing simple harmonic motion are represented as $\mathrm{y}_{1}=2 \sin (10 \mathrm{t}+\theta)$ and $\mathrm{y}_{2}=3 \cos 10 \mathrm{t} .$ The phase difference between the velocities of these waves is
  1. $\left(\theta+\frac{\pi}{2}\right)$
  2. $-\theta$
  3. $\left(\theta-\frac{\pi}{2}\right)$
  4. $\theta$

Solution

$\begin{aligned} \mathrm{y}_{1} &=2 \sin (10 \mathrm{t}+\theta) \quad \therefore \mathrm{V}_{1}=\frac{\mathrm{dy}_{1}}{\mathrm{dt}}=20 \cos (10 \mathrm{t}+\theta) \\ \mathrm{y}_{2} &=3 \cos 10 \mathrm{t} \\ &=-30 \cos \left(\frac{\pi}{2}-10 \mathrm{t}\right) \\ &=-30 \cos \left(10 \mathrm{t}-\frac{\pi}{2}\right) \\ &=30 \cos \left(10 \mathrm{t}-\frac{\pi}{2}+\pi\right) \\ &=30 \cos \left(10 \mathrm{t}+\frac{\pi}{2}\right) \end{aligned}$ $\therefore$ phase difference between $\mathrm{V}_{1}$ and $\mathrm{V}_{2}$ $=(10 t+\theta)-\left(10 t+\frac{\pi}{2}\right)=\left(\theta-\frac{\pi}{2}\right)$

Asked in: MHT CET 2020 (12 Oct Shift 2)

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