I have the basic set-up where you have a block attached to avertical wall via a
ID: 1755753 • Letter: I
Question
I have the basic set-up where you have a block attached to avertical wall via a spring. You are able to compress the blocktowards the wall on a frictionless table and let it go back andforth horizontally. The goal, given the mass of the block, amplitude of motion,and period / frequency of the motion, is to find the accelerationwhen it reaches the maximum displacement. This would usually be easy. Simply determine the springconstant, 'k', and set hooke's law equal to newton's law: -kx = ma. Here's the question:In most physics problems the mass of the spring is not considered.However, in my problem, I am given the mass of the spring which isonly slightly less than the mass of the block. How does havingmassive springs vs massless springs effect solving theproblem? I have the basic set-up where you have a block attached to avertical wall via a spring. You are able to compress the blocktowards the wall on a frictionless table and let it go back andforth horizontally. The goal, given the mass of the block, amplitude of motion,and period / frequency of the motion, is to find the accelerationwhen it reaches the maximum displacement. This would usually be easy. Simply determine the springconstant, 'k', and set hooke's law equal to newton's law: -kx = ma. Here's the question:
In most physics problems the mass of the spring is not considered.However, in my problem, I am given the mass of the spring which isonly slightly less than the mass of the block. How does havingmassive springs vs massless springs effect solving theproblem?
Explanation / Answer
In most physics problems the mass of the spring is not considered.However, in my problem, I am given the mass of the spring which isonly slightly less than the mass of the block. How does havingmassive springs vs massless springs effect solving theproblem? when the mass of the spring is negligible time period =T=2m/k where m is the mass attached toend of the spring . if ms is the mass of the spring then the same expression willbecome T=2[m+ms/3]/k similarly for other expressions also there will be a change whenthe mass of the spring is considered.