2. A simple harmonic oscillator consists of a 0.5-kg block attached to a spring

Question

2. A simple harmonic oscillator consists of a 0.5-kg block attached to a spring and vibrates about an equilibrium position at x -0. The force constant of the spring is 32 N/m and the block oscillates on a horizontal frictionless surface described by the general equation, x(t)-(0.30 m) cos (or + 0.85). (a) Determine the displacement of this block (x) at time t- 1.2 s. [-0.156 m] (b) What is the velocity of this object at timet-1.2 s? [2.1 m/s, positive x-direction] (c) At what position (x) from the equilibrium position is the kinetic energy of the block equal to half its maximum value? [0.212 m]

Explanation / Answer

here,

mass , m = 0.5 kg

spring constant , K = 32 N/m

angular frequency , w = sqrt(k/m)

w = sqrt(32/0.5) = 8 rad/s

x(t) = 0.3 * cos(w*t + 0.85 )

x(t) = 0.3 * cos(8*t + 0.85 ) ....(1)

a)

at t = 1.2 s

x(1.2) = 0.3 * cos(8*1.2 + 0.85 ) = - 0.156 m

b)

differentiating equation (1) for v

v(t) = - 2.4 * sin(8t + 0.85)

at t = 1.2 s

v(t) = - 2.4 * sin(8 * 1.2 + 0.85)

v(t) = 2.1 m/s

c)

the maximum kinetic energy = maximum potential energy = 0.5 * k * A^2 = 0.5 * 32 * 0.3^2 = 1.44 J

let the kinetic energy is at half of its maximum value at x

using conservation of energy

KEm = KEm/2 + 0.5 * K * x^2

1.44/2 = 0.5 * 32 * x^2

solving for x

x = 0.212 m

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