A 0.12 kg block is attached to the end of a horizontal spring. The spring, which

Question

A 0.12 kg block is attached to the end of a horizontal spring. The spring, which is initially in equilibrium gets compressed. The compressed spring-block system has an elastic potential energy of 1.8 J. The spring is released and starts oscillating back and forth with a frequency of 3 Hz. Find a) the spring constant b) the amplitude of oscillation c) the maximum speed of the block and where it occurs d) the maximum acceleration of the block and where it occurs e) the speed of the block when it passes at a distance x = 0.2 m away from the initial equilibrium position

Explanation / Answer

1) Frequency F = 1/2 pi [ sqrt [ k/M ]

squaring and derive for k = (2 pi F)^2 * M = (2*3.14*3)^2 (0.12) = 42.34 N/m

b) Energy conservation (1/2) k A^2 = 1.8 J ,

derving for amplitude A = sqrt [ 2*1.8 /K ] = 0.29m or 29cm

c) maximum speed is at equilibrium

V= A (2 pi F ) = 0.29 * (2*3.14*3) = 5.48m/s  

d) Acceleration is maximum at the extremes so

a = - A w^2 = -A (2 pi F )^2=- 0.29 * (2*3.14*3)^2 = -103m/s^2

Negative sign indicates that acceleration is opposite to displacement

e) conservation of energy

(1/2) k A^2 = (1/2) kx^2 + (1/2) m V^2

1.8 = (1/2)(42.34)(0.2)^2 + (1/2) (0.12)V^2

V = 3.98m/s

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