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Unformatted text preview: Physics 105 Solutions to Problem Set 4 By Stephen Glassman, Fall 2007 5.1 We actually did this exact problem in the last problem set. That previous solution (slightly edited) reappears below. (a) The mass will be in an equilibrium position when the net force on it vanishes. Therefore, we insist that at x , kx mg = 0. We conclude that x = mg k . Now we can express the force at arbitrary x as F ( x ) = kx mg = kx + kx = k ( x x ). (b) Recall that gravitational potential energy is related to height as Δ PE g = mg Δ h and that spring potential is related to displacement from the natural length as Δ PE s = 1 2 kx 2 . The total potential is then PE = mgx + 1 2 kx 2 . Let U ~a ( ~ b ) de note the potential energy at ~ b relative to ~a . Then by equation (4.16), which implies that U x ( x ) = U ( x ) U ( x ), we have: U x ( x ) = mgx + 1 2 kx 2 mgx 1 2 kx 2 = kx x + 1 2 kx 2 + kx 2 1 2 kx 2 = kx x + 1 2 kx 2 + 1 2 kx 2 = 1 2 k ( x x ) 2 where we used mg = kx from our calculation of the equilibrium point above. 5.2 We begin with the sketch. We rewrite our equation as: U A = h ( e R/S r/S 1 ) 2 1 i Up to scaling factors for the axes, there is one free parameter, R/S . We’ll see shortly that this parameter sets the location of the minimum. We’ll choose R/S = 1 for our graph below. To find the equilibrium separation, we could resort to standard calculus techniques. Alternatively, note that the only nonconstant piece is ( e ( R r ) /S 1 ) 2 which is strictly nonnegative and vanishes if and only if r = R . So we set r = R + x . Then U ( x ) = A h ( e x/S 1 ) 2 1 i . Now we use e x/S ≈ 1+ x 2 S 2 to derive U ( x ) ≈ A x 2 S 2 1 . 1 1 . 5 . 5 1 1 . 5 2 1 2 3 4 5 U/A r/S Figure 1: U ( r ) This has the form U = const + 1 2 kx 2 for const = A and k = 2 A S 2 . Alternatively, we could calculate U ( x ), U ( x ), and U 00 ( x ) explicitly and evaluate the Taylor expansion to second order directly. The answer is the same. 5.6 We are asked for a solution of the form x ( t ) = A cos ( ωt δ ), where A and δ are unknown. Fortunately, we are also given two equations. First, we are told that x max = 2 x . Of course, x ( t ) will be maximized when cos ( ωt δ ) = 1 which implies that x ( t max ) = A . Therefore, A = 2 x . Second, we are told that x (0) = x which implies that A cos δ = x . Then cos δ = 1 2 and δ = ± π 3 . Our expression x ( t ) becomes x ( t ) = 2 x cos ( ωt ± π 3 ) . We calculate ˙ x ( t ) = 2 x ω sin ( ωt ± π 3 ) and ˙ x (0) = ∓ x ω √ 3 . We know the mass was kicked with negative velocity, so we need the upper sign. Our final expression is then: x ( t ) = 2 x cos ( ωt + π 3 ) ....
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This note was uploaded on 04/01/2008 for the course PHYSICS 105 taught by Professor Edgarknobloch during the Fall '07 term at Berkeley.
 Fall '07
 EdgarKnobloch
 Mass, Light

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