1. A 2 kg block is attached to a spring and placed on a horizontal smooth surface. A horizontal force of 20 N is required to hold the block at rest when it is pulled 0.2 m from its equilibrium position. The block is released from its rest from this point, and is subsequently undergoes simple harmonic motion. Find a) the spring stiffness; b) the frequency of oscillations; c) the maximum speed of the block. Where does this maximum speed occur? Find the total energy of the oscillating system. Find the speed when the position equals one third of the maximum value.
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Force F = 20 N
mass m = 2 kg
maximum displacement from the mean position x_max = 0.2 m
Because, F = k*x
=> stiffness constant k = F/x = 20/0.2 = 100 N/m --Answer
b.) Frequency of oscillation f = (1/2*pi)*sqrt(k/m) = (1/2*pi)*sqrt(100/2) = 1.125 Hz --Answer
c.) By energy law,
Total energy = potential energy at maximum displacement = maximum K.E. at 0 displacement from mean
(1/2)*k*x_max^2 = (1/2)*m*v_max^2
=> v_max = sqrt(k/m)*x_max = sqrt(100/2)*0.2 = 1.414 m/s ...
This solution is provided in 376 words. It includes calculations for spring stiffness, and uses energy law to calculate total energy. Many other simple calculations are done, and an attached .jpg file is given with a diagram.