An object of mass m = 8.0 kg is attached to an ideal spring and allowed to hang in the earth's gravitational field. The spring stretches $2.2 \mathrm {~cm}$ before it reaches its equilibrium position. If it were now allowed to oscillate by this spring, what would be its frequency? A) 3.4 Hz B) 0.28 x 10^-3 Hz C) 0.52 Hz D) 1.6 Hz

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Quizplus · Accepted Answer

The frequency of an object attached to an ideal spring is given by:$$f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$$where k is the spring constant and m is the mass of the object.In this case, we are given that the object has a mass of 8.0 kg and that the spring stretches 2.2 cm before it reaches its equilibrium position. We can use this information to find the spring constant:$$k = \frac{mg}{x}$$where g is the acceleration due to gravity (9.8 m/s^2) and x is the displacement of the spring (0.022 m).Plugging in these values, we get:$$k = \frac{(8.0 	ext{ kg})(9.8 	ext{ m/s}^2)}{0.022 	ext{ m}} = 3528 	ext{ N/m}$$Now we can plug in the values for k and m into the equation for frequency:$$f = \frac{1}{2\pi}\sqrt{\frac{3528 	ext{ N/m}}{8.0 	ext{ kg}}} = 3.4 	ext{ Hz}$$Therefore, the correct answer is A.

An Object of Mass M = 8 2.2 cm2.2 \mathrm {~cm}2.2 cm Before It Reaches Its Equilibrium Position

An Object of Mass M = 8 $2.2 \mathrm {~cm}$ Before It Reaches Its Equilibrium Position