InterviewSolution
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InterviewSolution
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A particle is subjected to two simple harmonic motions `x_1=A_1 sinomegat` `and x_2=A_2sin(omegat+pi/3)` Find a the displacement at t=0, b. the maxmum speed of the particle and c. the maximum acceleration of the particle |
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Answer» a. At t=0, `x_1=A_1sinomegat=0` `and x_2=A_2sin(omegat+pi/3)` `=A+2sin(pi/3)=(A_2/sqrt3)/2` Thus the resultant displacement at t=0 is `x=x_1+x_2=(A_2sqrt3)/2` b. The resultant of the two motions is a simple harmonic motion of the same angular frequency omega. The amplitude of the resultant motion is `A=sqrt(A_1^2+A_2^2+2A_1A_2cos(pi/3))` `=sqrt(A_1^2+A_2^2+A_1A_2)` the maximum speed is `=v_(max)=Aomega=omegasqrt(A_1^2+A_2^2+A_1A_2)` c. The maximum acceleration is `a_(ma)=Aomega^2=omega^2 sqrt(A_1^2+A_2^2+A_1A_2)` |
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