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InterviewSolution
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A particle is subjected to two simple harmonic motions of sae time period in the same direction. The amplitude of the first motion is 3.0 cm and that of the second is 4.0 cm. Find the resultant amplitude if the phase difference between the motion is a. `0^@, b. 60^@, c. 90^@` |
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Answer» Correct Answer - A: 7cm B: 6.1cm C: 5cm The particle is subjected to two SHMs of same time period in the same direction. Given `r_1=3`cm, `r_2=4`cm and `phi`=phase difference Resultant amplitude `R=sqrt(r_1^2+r_2^2+2r_1r_2cos phi)` a. when `phi=0^@` `R=sqrt((3^2+4^2+2.3.4cos0^/2))` `=sqrt49=7cm` b. when `phi=60^@` `R=sqrt((3^2+4^2+2.3.4cos60^@))` `=sqrt37=6.1cm` c. when `phi=90^@` `R=sqrt((3^2+4^2+2.3.4cos90^@))` `=sqrt25=5cm` |
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