An object of mass 8.0 kg is attached to an ideal massless spring and allowed to hang in the Earth's gravitational field.The spring stretches 3.6 cm before it reaches its equilibrium position.If this system is allowed to oscillate,what will be its frequency?
A) 2.6 Hz
B) 0.0045 Hz
C) 0.67 Hz
D) 2.1 Hz

Question

Quizplus · Accepted Answer

The frequency of an oscillating spring is given by f = 1/2π√(m/k), where m is the mass of the object and k is the spring constant. In this case, m = 8.0 kg and the spring stretches 3.6 cm = 0.036 m before it reaches its equilibrium position. The spring constant is given by k = mg/x, where g is the acceleration due to gravity (9.8 m/s^2). Therefore, k = (8.0 kg)(9.8 m/s^2)/(0.036 m) = 2157.14 N/m. Substituting these values into the equation for frequency, we get f = 1/2π√(8.0 kg/2157.14 N/m) = 2.6 Hz.

An Object of Mass 8