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# Answer to Question #102596 in Mechanics | Relativity for Ojugbele Daniel

Question #102596
The position of a simple harmonic oscillator as a function of time is given by x=4.6cos(7πt/6 + π/8) where t is in second and x is in meter. Find: (i) the frequency (ii) the velocity and acceleration at t=0.
1
2020-02-10T09:27:04-0500

As per the question,

The equation of the wave "x=4.6\\cos(\\dfrac{7\u03c0t}{6} + \u03c0\/8)"

Now comparing this the general SHM equation, "X=A cos(\\omega t+\\phi)"

a)

"\\omega=\\dfrac{7\\pi}{6}"

"\\Rightarrow 2\\pi f=\\dfrac{7\\pi}{6}"

"\\Rightarrow f=\\dfrac{7\\pi}{6\\times2\\pi}=\\dfrac{7}{12}Hz"

b)

"x=4.6\\cos(\\dfrac{7\u03c0t}{6} + \u03c0\/8)"

"v=\\dfrac{dx}{dt}=-4.6\\times \\dfrac{7\\pi}{6}\\sin(\\dfrac{7\\pi t}{6}+\\dfrac{\\pi}{8})"

at t=0

"v=-4.6\\times \\dfrac{7\\pi}{6}\\sin(\\dfrac{\\pi}{8})=-11.91m\/sec"

"a=\\dfrac{dv}{dt}=\\dfrac{d^2x}{dt^2}=-4.6\\times \\dfrac{7\\pi}{6}\\times \\dfrac{7\\pi}{6}\\cos(\\dfrac{7\\pi t}{6}+\\dfrac{\\pi}{8})"

at t=0,

"a=-4.6\\times \\dfrac{7\\pi}{6}\\times \\dfrac{7\\pi}{6}\\sin(\\dfrac{\\pi}{8})=-\\dfrac{225.4\\pi^2}{36}\\times\\dfrac{1}{\\sqrt{2}}"

"a=-43.73m\/sec^2"

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