Answer to Question #217836 in Mechanical Engineering for john Smith

Question #217836




A point moves in a straight line with simple harmonic motion. When it is 500mm to the right of the mean position and moving to the left, the velocity is 3m/s. When it is 1 meter to the left of the mean position and moving to the right, the velocity is 2m/s. Determine the amplitude.


1
Expert's answer
2021-07-17T04:52:05-0400

"tan\\phi = \\frac{v(0)}{\\omega x(0)} = \\frac{3}{2\\times0.5}"


"\\phi = 71.56^0"


Amplitude :


"x_m = \\frac{x(0)}{cos\\phi} = \\frac{1}{cos(71.56)} =3.16 m = 316 cm"




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