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# Simple Harmonic Motion defined by a reference circle from which come five SHM equations

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Point P is moving in a circle at constant speed. On a diameter on the x axis, point Q moves in such a way that the x coordinates of both P and Q remain the same.
SEE ATTACHMENT #1 for a diagram and explanation of parameters.
PART a. First with parameters then with numbers, express x as a function of time.
PART b. Since the above function represents position x, its derivative would be the rate of change of position, namely the velocity. Take the derivative to write v(t).
PART c. Similarly, the derivative of v(t) is the acceleration, a(t). Take the derivative to express a(t).
PART d. Note that in a(t), the right side of x(t) can replace all parameters past w^2. Use this to eliminate t and write a(x).
PART e. Use the identity '(sin b)^2 + (cos b)^2 = 1' to eliminate t from v(t) and express the result as v(x).
NOTE; The above five equations represent any SHM in which the motion starts when the moving point Q is at xm at t=0. The basic first step is to write x(t) from stated conditons, then use derivatives, etc., to obtain the other equations.
If the initial starting point of Q is other than xm, SEE ATTACHMENT #2 for an example showing x(t) for that situation.
PART f. If known parameters are: xm= .24 m and angular velocity w= 4 rad/sec, use your numerical equations to calculate the following;
period T, x at.25 sec=, v at .25 sec, a at .25 sec, v at -.18 m, and a at -.18 m.

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#### Solution Preview

PART a. From the triangle formed by base x and hypotenuse xm with included angle wt, the equation giving x(t) is:
(1) x = xm cos wt = .24 cos (4t).
PART b. Taking the derivative ...

#### Solution Summary

Simple Harmonic Motion is expressed.

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