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[Smilebuffalo]

A particle, whose displacement is x, moves in simple harmonic motion.
Find x as a function of t if:

acceleration = −4x

and if x = 3 and v = −6root3 when t = 0.

 

[scardizzle]

intergral of a = 1/2v^2

therefore Intergral -4x = -2x^2 + c = 1/2 v^2

therefore v^2 = -4x^2 + c

when x = 3 v = -6root 3

therefore 108 = -4 x 9 + c

therefore c = 144

v^2 = -4x^2 + 144

therefore v = -(-4x^2 + 144)^1/2

therefore dx/dt = -(144 - 4x^2)^1/2

therefore dt/dx = -1/(144 - 4x^2)^1/2


therefore t = 1/2 cos^-1(x/6) + c

when t = 0 x = 3

therefore 0 = pi/6 + c

therefore c = - pi/6

making x the subject

t + pi/6 = 1/2cos^-1(x/6)

cos(2t + pi/3) = x/6

therefore x= 6cos(2t + pi/3)