InterviewSolution
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InterviewSolution
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A particle is subjected to two simple harmonic motions. `x_(1) = 4.0 sin (100pi t)` and `x_(2) = 3.0 sin(100pi t + (pi)/(3))` Find (a) the displacement at `t = 0` (b) the maximum speed of the particle and (c ) the maximum acceleration of the particle. |
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Answer» (a) At `t = 0, x_(1) = A_(1) sin omegat = 0` and `x_(2) = A_(2) sin (omegat + pi//3)` `= A_(2) sin (pi//3) = (A_(2)sqrt(3))/(2)` Thus, the resultant displacement at `t = 0` is `x = x_(1) + x_(2) = A_(2) sqrt(3)/(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_(1)A_(2) cos(pi//3)), = sqrt(A_(1)^(2) + A_(2)^(2) + A_(1)A_(2))`. The maximum speed is `u_("mass") = Aomega = omegasqrt(A_(1)^(2) + A_(2)^(2) + A_(1)A_(2))` (c) The maximum acceleration is `a_(max) = Aomega^(2) = omega^(2) sqrt(A_(1)^(2) + A_(2)^(2) + A_(1)^(2))`. |
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