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MHT CET · Physics · Oscillations

Two S.H.Ms. are represented by equations \(\mathrm{y}_1=0.1 \sin \left(100 \pi \mathrm{t}+\frac{\pi}{3}\right)\) and \(\mathrm{y}_2=0.1 \cos (100 \pi \mathrm{t})\) The phase difference between the speeds of the two particles is

  1. A \(\frac{\pi}{3}\)
  2. B \(-\frac{\pi}{6}\)
  3. C \(+\frac{\pi}{6}\)
  4. D \(-\frac{\pi}{3}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(-\frac{\pi}{6}\)

Step-by-step Solution

Detailed explanation

Given: \(y_1=0.1 \sin \left(100 \pi t+\frac{\pi}{3}\right)\) and
\(\begin{aligned}
\mathrm{y}_2= & 0.1 \cos (100 \pi \mathrm{t}) \\
\therefore \quad \mathrm{v}_1=\frac{\mathrm{dy}_1}{\mathrm{dt}} & =0.1 \times 100 \pi \cos \left(100 \pi \mathrm{t}+\frac{\pi}{3}\right) \\
\mathrm{v}_2=\frac{\mathrm{dy}_2}{\mathrm{dt}} & =-0.1 \times 100 \pi \sin (100 \pi \mathrm{t}) \\
& =(0.1 \times 100 \pi) \cos \left(100 \pi \mathrm{t}+\frac{\pi}{2}\right)
\end{aligned}\)
Phase difference of velocity of first particle with respect to the velocity of second particle at \(t=0\) is
\(\therefore \quad \Delta \phi=\phi_1-\phi_2=\frac{\pi}{3}-\frac{\pi}{2}=\frac{-\pi}{6}\)