InterviewSolution
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InterviewSolution
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The diplacement of a particle varies with time according to the relation `y=a"sin"omegat+b " cos"omegat`.A. The motion is oscillatory but not SHMB. The motion is SHM with amplitude a+bC. The motion is SHM with amplitude `a^(2)+b^(2)`D. The motion is SHM with amplitude `sqrt(a^(2)+b^(2))` |
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Answer» Correct Answer - D (d) Accoding to the question , the displacement `y=a "sin"omegat+b "cos"omegat` Let `a=A"sin"phi " and "b=A"cos" phi` Now , `a^(2)+b^(2)=A^(2)"sin"^(2)phi+A^(2)"cos"^(2)phi` `=A^(2)impliesA=sqrt(a^(2)+b^(2))` `therefore y=A"sin"phi, "sin" omegat+A"cos"phi."cos"omegat` `=A"sin"(omegat+phi)` `implies(dy)/(dt)=Aomega"cos"(omegat+phi)` `implies(d^(2)y)/(dt^(2))=-Aomega^(2)"sin"(omegat+phi)=-Ayomega^(2)=(-Aomega^(2))y` `implies(d^(2)y)/(dt^(2))prop(-y)` Hence, it is an equation of SHM with amplitude, `A=sqrt(a^(2)+b^(2))`. |
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