Question 141: Oscillations

Question

Two simple harmonic motions are represented by the equations ${y_1} = 0.1\,\sin \left( {100\pi t + {\pi  \over 3}} \right)$ and ${y_2} = 0.1\,\cos \,\pi t.$ The phase difference of the velocity of particle $1$ with respect to the velocity of particle $2$ is

Accepted Answer

${v_1} = {{d{y_1}} \over {dt}} = 0.1 	imes 100\pi \cos \left( {100\pi t + {\pi  \over 3}} ight)$
${v_2} = {{d{y_2}} \over {dt}} =  - 0.1\pi sin\pi t = 0.1\pi cos\left( {\pi t + {\pi  \over 2}} ight)$
$	herefore$ Phase diff. $ = {\phi _1} - {\phi _2} = {\pi  \over 3} - {\pi  \over 2} = {{2\pi  - 3\pi } \over 6} = {\pi  \over 6}$

Answer

${\pi  \over 3}$

Answer

${{ - \pi } \over 6}$

Answer

${\pi  \over 6}$

Answer

${{ - \pi } \over 3}$

Oscillations

Two simple harmonic motions are represented by the equations {y_1} = 0.1\,\sin \left( {100\pi t + {\pi \over 3}} \right) and {y_2} = 0.1\,\cos \,\pi t. The phase difference of the velocity of particle 1 with respect to the velocity of particle 2 is

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