Question
The displacement of two particles executing simple harmonic motion are represented by equations, $\text{y}=4\sin(10\text{t}+\theta)$ and $\text{y}=5\cos10\text{t}$ What is the phase difference between the velocities of these particles?
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Answer
For 1st particle$\text{y}_1=4\sin(10\text{t}+\theta);$
Velocity $\frac{\text{dy}_1}{\text{dt}}$$=4\times10\cos(10\text{t}+\theta)$
$=40\cos(10\text{t}+\theta)$ For second particle $\text{y}_2=5\cos10\text{t}=5\sin(10\text{t}+\frac{\pi}{2})$
Velocity $\frac{\text{dy}_2}{\text{dt}}$$=5\cos\times10\cos(10\text{t}+\frac{\pi}{2})$
$=5\cos(10\text{t}+\frac{\pi}{2})$
Phase difference between velocities $=(10\text{t}+\theta)-(10\text{t}+\frac{\pi}{2})$
$=\Big(\theta-\frac{\pi}{2}\Big)$
Velocity $\frac{\text{dy}_1}{\text{dt}}$$=4\times10\cos(10\text{t}+\theta)$
$=40\cos(10\text{t}+\theta)$ For second particle $\text{y}_2=5\cos10\text{t}=5\sin(10\text{t}+\frac{\pi}{2})$
Velocity $\frac{\text{dy}_2}{\text{dt}}$$=5\cos\times10\cos(10\text{t}+\frac{\pi}{2})$
$=5\cos(10\text{t}+\frac{\pi}{2})$
Phase difference between velocities $=(10\text{t}+\theta)-(10\text{t}+\frac{\pi}{2})$
$=\Big(\theta-\frac{\pi}{2}\Big)$
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