A block of mass 0.585 kg is placed on the end of a spring which has a stiffness of 294 N/m. It is pulled down 2.80 cm and released. What is the frequency of oscillation?

angular freq=sqrt(k/m)=2PI*freq

you are given m, and k, solve for frequency

To find the frequency of oscillation, we need to use Hooke's Law and the equation for the period of a mass-spring system.

1. Start by finding the spring constant (k) using Hooke's Law equation: F = -kx, where F is the force applied on the spring, k is the spring constant, and x is the displacement from the equilibrium position.
In this case, the force applied on the spring is the weight of the block, which is given by F = mg, where m is the mass of the block and g is the acceleration due to gravity.
Therefore, F = (0.585 kg)(9.8 m/s^2) = 5.733 N.
Using Hooke's Law, 5.733 N = -k(0.028 m), since the block is pulled down 2.8 cm and the displacement is considered negative.
Solving for k, we get k = -5.733 N / -0.028 m = 204.75 N/m.

2. Now that we have the spring constant, we can calculate the frequency of oscillation using the equation: f = (1 / 2π) * √(k / m), where f is the frequency, k is the spring constant, and m is the mass of the block.
Plugging in the values, f = (1 / 2π) * √(204.75 N/m / 0.585 kg).
Simplifying, f = 3.86 Hz (rounded to two decimal places).

Therefore, the frequency of oscillation is approximately 3.86 Hz.