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Given x(t) for a certain SHM, find x, v, and a at a given instant.

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For a certain point executing SHM on an x axis with amplitude .40 m, the equation giving its location as a function of time is given by:
(1) x = .4 m cos (6 t)
a. At time t= .35 sec, find the location, the velocity and the acceleration of the moving point.
b. Find the first time after t=0 that the moving point is located at x= X1= -.31 m.

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Solution Summary

A function of simple harmonic motion is given. The location, velocity and acceleration is given.

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Part a. Step 1.
Substituting .35 sec for t in equation (1) should give you:
ANSWER: (2) x(.35) = - .20 m.
Step 2.
Since velocity v is dx/dt, taking the derivative ...

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