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2 August, 14:56

X = (7.1 m) cos[ (5πrad/s) t + π/5 rad] gives the simple harmonic motion of a body. at t = 1.6 s, what are the (a) displacement, (b) velocity, (c) acceleration, and (d) phase of the motion? also, what are the (e) frequency and (f) period of the motion?

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  1. 2 August, 15:18
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    Rather than answering your specific question I believe it's better to explain simple harmonic motion. We say some dynamical variable [tex]x[tex] is in simple harmonic motion if its value at time t is [tex]x (t) = A/cos (/omega t+/phi) [tex] for some constants [tex]A[tex] and [tex]/phi[tex]. The rate of change of [tex]x[tex] (in your case this is the velocity) as a function of time can be found by taking the derivative. In this case, [tex]v (t) = x' (t) = - A/omega/sin (/omega t + / phi) [tex]. The rate of change of the rate of change (in your case acceleration) can once again be found by taking the derivative [tex]a (t) = v' (t) = - A/omega^2/cos (/omega t + / phi) [tex]. The phase of the motion is just the argument of the cosine function, that is, [tex]/omega t+/phi[tex]. The frequency is the number of cycles which the motion makes in 1 second. This is of course just the inverse of the time it takes the oscillator in do one cycle. Since the cosine has period [tex]2/pi[tex] the time it take to make one cycle is [tex] T = / frac{2/pi}{/omega}[tex]. Therefore, the frequency is [tex]f=/frac{/omega}{2/pi}[tex].

    In your problem [tex] A=7.1m[tex], [tex]/omega=5/pi rad/s[tex] and [tex]/phi=/pi/5rad[tex]. You've just gotta replace this values in the equations above to solve for what the problem asks.
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