Stretch of Vertical Spring: Oscillating at 5.87 Hz

[yuvlevental]

An object, suspended from a vertical spring, oscillates up and down at 5.87 Hz. How much would this object stretch the spring, if it were hanging at rest?

 

[Doc Al]

How does the frequency of the simple harmonic motion relate to the spring constant and mass of the object?

 

[m2003]

The amount of stretch on the vertical spring would depend on the properties of the spring, such as its spring constant and length. However, we can calculate an estimate using Hooke's Law, which states that the force exerted by a spring is directly proportional to the distance it is stretched or compressed.

Assuming that the object is hanging at rest, the force of gravity acting on it would be equal to the force exerted by the spring. Therefore, we can set up the equation F = kx, where F is the force, k is the spring constant, and x is the distance the spring is stretched.

To find the distance x, we can rearrange the equation to x = F/k. The force F can be calculated using the formula F = mg, where m is the mass of the object and g is the acceleration due to gravity (9.8 m/s^2).

Substituting the values, we get x = (mg)/k. This means that the distance the spring is stretched is directly proportional to the mass of the object and inversely proportional to the spring constant. Therefore, the greater the mass of the object, the more the spring will stretch, and the smaller the spring constant, the more the spring will stretch.

In conclusion, the amount that the vertical spring will stretch when the object is hanging at rest will depend on the properties of the spring and the mass of the object. It is not possible to determine the exact amount without knowing these variables, but we can estimate it using Hooke's Law.