Question 12: Units , Dimensions and Error Analysis

Question

Two simple harmonic motions are represented by equations $y_1=0.5 \sin \left[200 \pi \mathrm{t}+\frac{\pi}{3}\right]$ and $y_1=0.5 \cos \pi \mathrm{t}$. The phase difference of the velocity of particle 1 with respect to the velocity of particle 2 is :

Accepted Answer

LetÃ¢ÂÂs first write the two SHMs (correcting the obvious typo so both have the same ÃÂ=200ÃÂ):
Ã¢ÂÂ¢ yÃ¢ÂÂ = 0.5Ã¢ÂÂsin(200ÃÂÃ¢ÂÂt + ÃÂ/3)
Ã¢ÂÂ¢ yÃ¢ÂÂ = 0.5Ã¢ÂÂcos(200ÃÂÃ¢ÂÂt)
Differentiate to get velocities:
$$v_1 = \frac{dy_1}{dt} = 0.5\,(200\pi)\cos(200\pi t + 	frac\pi3)              = 100\pi\;\cos\!\bigl(200\pi t + 	frac\pi3\bigr)$$
$$v_2 = \frac{dy_2}{dt} = 0.5\,( -200\pi)\,\sin(200\pi t)              = -100\pi\;\sin(200\pi t)$$
Rewrite $v_2$ in cosine form using $-\sin ÃÂ¸ = \cos(ÃÂ¸ + ÃÂ/2)$:
$$v_2 = 100\pi\;\cos\!\bigl(200\pi t + 	frac\pi2\bigr)$$
Now both velocities are cosines with phases
Ã¢ÂÂ¢ ÃÂÃ¢ÂÂ = ÃÂ/3
Ã¢ÂÂ¢ ÃÂÃ¢ÂÂ = ÃÂ/2
So the phase difference of vÃ¢ÂÂ w.r.t. vÃ¢ÂÂ is
$$\Delta ÃÂ = ÃÂÃ¢ÂÂ - ÃÂÃ¢ÂÂ = \frac\pi3 - \frac\pi2 = -\frac\pi6.$$
Answer:Ã¢ÂÂÃ¢ÂÂÃÂ/6 (Option C)

Answer

$\frac{\pi}{2}$

Answer

$\frac{\pi}{6}$

Answer

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

Answer

$\frac{-\pi}{2}$

Units , Dimensions and Error Analysis

Two simple harmonic motions are represented by equations y_1=0.5 \sin \left[200 \pi \mathrm{t}+\frac{\pi}{3}\right] and y_1=0.5 \cos \pi \mathrm{t}. The phase difference of the velocity of particle 1 with respect to the velocity of particle 2 is :

Choose the correct answer: