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?

Question

Quizplus · Accepted Answer

The correct answer is A because the frequency of an object attached to an ideal spring is given by f = (1/2π)√(k/m), where k is the spring constant and m is the mass of the object. In this case, we can find the spring constant by using the equation F = kx, where F is the force applied to the spring, x is the displacement of the spring, and k is the spring constant. The force applied to the spring is equal to the weight of the object, which is mg, where g is the acceleration due to gravity. The displacement of the spring is 2.2 cm, or 0.022 m. Substituting these values into the equation, we get k = mg/x = (8.0 kg)(9.8 m/s^2)/(0.022 m) = 352.7 N/m. Substituting this value of k into the equation for frequency, we get f = (1/2π)√(k/m) = (1/2π)√(352.7 N/m / 8.0 kg) = 3.4 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