A mass m = 8.0 kg is attached to a spring and allowed to hang in the Earth's gravitational field.The spring stretches 3.6 cm before it reaches its equilibrium position.If allowed to oscillate,what would be its frequency? A)2.6 Hz B)0.45 × 10^-3 Hz C)0.67 Hz D)2.1 Hz

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Quizplus · Accepted Answer

The correct answer is A because the frequency of an oscillating mass-spring system is given by f = 1/2π√(k/m), where k is the spring constant and m is the mass. The spring constant can be determined from the amount of stretch, which is 3.6 cm = 0.036 m. The force exerted by the spring is given by F = kx, where x is the displacement from the equilibrium position. At equilibrium, the force exerted by the spring is equal to the force of gravity, which is mg. Therefore, k = mg/x = (8.0 kg)(9.8 m/s^2)/(0.036 m) = 2157 N/m. The frequency is then f = 1/2π√(2157 N/m / 8.0 kg) = 2.6 Hz.

A Mass M = 8