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A 0.50-kg mass is attached to a spring of spring constant 20 N/m along a horizontal, frictionless surface. The object oscillates in simple harmonic motion and has a speed of 1.5 m/s at the equilibrium position. At what loctaion are the kinectic energy and the potential energy the same?

Added Mon, 03 Aug '15

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Vmax = wA

Amplitude A = 1.5/w = 1.5*sqrt(0.5/20) = 0.237 m

kinetic and potential energies are equal say at point x

0.5mv^2 =0.5kx^2 (since v^2 = w^2[A^2-x^2]=k(A^2-x^2)/m )

0.5m*k(A^2-x^2)/m = 0.5kx^2

(A^2-x^2) = x^2

2x^2 = A^2

x = A *sqrt(1/2) = sqrt(0.237/2) = 0.167 m

Added Mon, 03 Aug '15

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