Question

Two simple harmonic motions are represented by the equations y₁ = 0.1 sin $\left( 100\pi t + \frac{\pi}{3} \right)$and y₂ = 0.1 cospt. The phase difference of the velocity of particle 1 with respect to the velocity of particle 2 is –

• A. $\frac{\pi}{6}$
• B. $\frac{- \pi}{3}$
• C. $\frac{\pi}{3}$
• D. $\frac{- \pi}{6}$

Answer 1

Answer: (D)

$\frac{- \pi}{6}$

Answer 2

y₂ = 0.1 cos pt = 0.1 sin [p/2 + pt]

V₁ = $\frac{dy_{1}}{dt}$ = 0.1 × 100p cos [100pt + p/3]

V₂ = p × 0.1 cos [p/2 + pt]

so phase difference of (1) w.r.t. to (2)

= $\frac{\pi}{3}$ – $\frac{\pi}{2}$ = –$\frac{\pi}{6}$