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

Question

Quizplus · Accepted Answer

The correct answer is A because the frequency of an object attached to an ideal spring is given by:$$f = \sqrt{\frac{k}{m}}$$where k is the spring constant and m is the mass of the object. In this case, the spring constant can be calculated from the amount of stretch:$$k = \frac{mg}{x}$$where g is the acceleration due to gravity and x is the amount of stretch. Substituting these values into the equation for frequency, we get:$$f = \sqrt{\frac{mg}{mx}} = \sqrt{\frac{g}{x}}$$Plugging in the given values, we get:$$f = \sqrt{\frac{9.81 \mathrm{~m/s^2}}{0.022 \mathrm{~m}}} = 3.4 \mathrm{~Hz}$$

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