2. When a 625g mass is attached to a vertical spring, the spring stretches by 16cm. How much mass must be attached to the spring to result in a .78s period of oscillation?

To solve this problem, we need to use Hooke's Law for springs and the formula for the period of oscillation.

1. Hooke's Law for springs states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position. Mathematically, this can be represented as:

F = -kx

where F is the force exerted by the spring, k is the spring constant, and x is the displacement from the equilibrium position.

2. The formula for the period of oscillation (T) of a mass-spring system is given by:

T = 2π√(m/k)

where T is the period, m is the mass attached to the spring, and k is the spring constant.

3. In this problem, we are given that a 625g mass (0.625 kg) attached to the spring causes a 16 cm (0.16 m) displacement. We need to find the mass that will result in a period of 0.78 s.

4. First, we need to determine the spring constant (k) using the given information. We can rearrange Hooke's Law to solve for k:

F = -kx
k = -F/x

The force exerted by the spring can be found using the weight formula:

F = mg

Substituting the given values:

F = (0.625 kg)(9.8 m/s^2) = 6.125 N

Finally, substituting the values in the formula for k:

k = -(6.125 N)/(0.16 m) = -38.28125 N/m (rounding to two decimal places)

5. Now that we have the spring constant (k), we can rearrange the formula for the period to solve for the mass (m):

T = 2π√(m/k)
T^2 = 4π^2(m/k)
m = (T^2)(k)/(4π^2)

Substituting the given values:

m = (0.78 s)^2(-38.28125 N/m)/(4π^2)

Calculating this expression will give us the mass required to result in a period of 0.78 s.

Please note that the negative sign in calculations arises from the definition of the spring force direction, which can be either compression or elongation.

k=mg/x=0.625•9.8/0.16 =38.3 N/m

T=2•π•sqrt(m/k)
m= k•T²/4π²