a) The force exerted by a one-dimensional spring, fixed at | StudySoup

Textbook Solutions for Classical Mechanics

Chapter 4 Problem 4.9

Question

(a) The force exerted by a one-dimensional spring, fixed at one end, is \(F = -kx\), where \(x\) is the displacement of the other end from its equilibrium position. Assuming that this force is conservative (which it is) show that the corresponding potential energy is \(U=\frac{1}{2}kx^2\), if we choose \(U\) to be zero at the equilibrium position. (b) Suppose that this spring is hung vertically from the ceiling with a mass \(m\) suspended from the other end and constrained to move in the vertical direction only. Find the extension \(x_o\) of the new equilibrium position with the suspended mass. Show that the total potential energy (spring plus gravity) has the same form \(\frac{1}{2}ky^2\) if we use the coordinate \(y\) equal to the displacement measured from the new equilibrium position at \(x=x_{\mathrm{o}}\) (and redefine our reference point so that \(U = 0\) at \(y = 0\)).

Solution

Step 1 of 6

(a)

The expression can be given as,

                                                               

Subscribe to view the
full solution

Title Classical Mechanics 0 
Author John R Taylor
ISBN 9781891389221

a) The force exerted by a one-dimensional spring, fixed at

Chapter 4 textbook questions

×

Login

Organize all study tools for free

Or continue with
×

Register

Sign up for access to all content on our site!

Or continue with

Or login if you already have an account

×

Reset password

If you have an active account we’ll send you an e-mail for password recovery

Or login if you have your password back