Spring calculations: A spring stretches 0.150n m when a 0.3 kgmass is gently suspended from it. The spring is then set uphorizontally with the 0.3 kg mass resting on the frictionless table.The mass is pulled so that the spring is stretched 0.1 m from theequilibrium point and released from rest. Determine a) the springstiffness constant k, b) the amplitude of the maximum acceleration.
Question
Spring calculations: A spring stretches 0.150n m when a 0.3 kgmass is gently suspended from it. The spring is then set uphorizontally with the 0.3 kg mass resting on the frictionless table.The mass is pulled so that the spring is stretched 0.1 m from theequilibrium point and released from rest. Determine a) the springstiffness constant k, b) the amplitude of the maximum acceleration.
Solution
a) The spring stiffness constant k can be determined using Hooke's Law, which states that the force exerted by a spring is proportional to the displacement from its equilibrium position. The formula is F = kx, where F is the force, k is the spring constant, and x is the displacement.
In this case, the force is the weight of the 0.3 kg mass, which is F = mg = 0.3 kg * 9.8 m/s^2 = 2.94 N. The displacement x is 0.150 m. So we can solve for k:
2.94 N = k * 0.150 m k = 2.94 N / 0.150 m = 19.6 N/m
b) The amplitude of the maximum acceleration can be found using the formula for the acceleration of a mass-spring system, a = -kx/m, where a is the acceleration, k is the spring constant, x is the displacement, and m is the mass.
In this case, the displacement x is 0.1 m (the distance the spring is stretched from the equilibrium point), the spring constant k is 19.6 N/m, and the mass m is 0.3 kg. So we can solve for a:
a = -19.6 N/m * 0.1 m / 0.3 kg = -6.53 m/s^2
The negative sign indicates that the acceleration is directed towards the equilibrium position, as expected. The amplitude of the maximum acceleration is the absolute value, so it's 6.53 m/s^2.
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