# Physics 16 Final

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Physics 16 Final Exam Take-Home September 24, 2006 1a. L = T − V , where L is the Langragian, T is the kinetic energy, and V is the potential energy. m (x0 (t)2 +y 0 (t)2 ) T = 2  V = k −d +  2 k −d + q x(t)2 + (−` + y(t))2 L = − k −d + 2 k −d + q 2 2 2  +k −d + (−` + x(t)) + y(t) q   2 2  + 2 k −d + q 2 2 2 2 x(t) + (−` + y(t))  −2 k −d + 2 2 (` + x(t)) + y(t) q x(t)2 + (` + y(t))2 2 ! (−` + x(t)) + y(t) q  −k −d + q 2 + 2 q (` + x(t))2 +
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Physics 16 Final Exam Take-Home September 24, 2006 1a. L  =  T  − V   , where  L  is the Langragian,  T   is the kinetic energy, and  V   is thepotential energy. T   =  m ( x  ( t ) 2 + y  ( t ) 2 ) 2 V   =  k  − d  +   ( −   +  x ( t )) 2 +  y ( t ) 2  2 + k  − d  +   (   +  x ( t )) 2 +  y ( t ) 2  2 +2 k  − d  +   x ( t ) 2 + ( −   +  y ( t )) 2  2 + 2 k  − d  +   x ( t ) 2 + (   +  y ( t )) 2  2 L  = −  k  − d  +   ( −   +  x ( t )) 2 +  y ( t ) 2  2  − k  − d  +   (   +  x ( t )) 2 +  y ( t ) 2  2 − 2 k  − d  +   x ( t ) 2 + ( −   +  y ( t )) 2  2 − 2 k  − d  +   x ( t ) 2 + (   +  y ( t )) 2  2 + m ( x  ( t ) 2 + y  ( t ) 2 ) 2 To ﬁnd the energy,  E  , we can use E   = N   i =1 ∂L∂   ˙ q  i ˙ q  i − L where the  q  i ’s are the variables in the Langrangian. E   = ( mx  ( t ) 2 − my  ( t ) 2 ) −  m 2  ( x  ( t ) 2 +  y  ( t ) 2 ) +  V E   =  T   +  V  b. If   d  = 0, then  L  simpliﬁes to L  = 12( x  ( t ) 2 +  y  ( t ) 2 ) − 6 k ( x ( t ) 2 +  y ( t ) 2 +   2 )This has symmetry under the inﬁnitesimal transformation  x  →  x  + y,y  → y − x  to ﬁrst order in   . Thus, by Noether’s theorem, the followingquantity is conserved:1   i ∂L∂   ˙ q  i K  i ( q  ) =  m (˙ xy −  ˙ yx )c. This is easiest if we just write down the equations of motion using  f   = ma  and then make some approximations using Taylor expansions. Summingthe force contributions from each spring, we get: m ¨ x  = 2 k  (  − x )  1 −  d   (  − x ) 2 +  y 2  − (   +  x )  1 −  d   (   +  x ) 2 +  y 2  − 4 k  x  1 −  d   (  − y ) 2 +  x 2  +  x  1 −  d   (   +  y ) 2 +  x 2  +  F  0  cos( ωt ) m ¨ y  = − 4 k  (  − y )  1 −  d   (  − y ) 2 +  x 2  − (   +  y )  1 −  d   (   +  y ) 2 +  x 2  2 k  y  1 −  d   (  − x ) 2 +  y 2  +  y  1 −  d   (   +  x ) 2 +  y 2  +  F  0  cos( ωt )Since we are given that the oscillations about the equilibrium positionare small, we can Taylor expand both  x  and  y  about zero, using the followingidentities: x   x 2 + (  − y ) 2  ≈  x 1 − y  ≈  x  +  xy 2 x   x 2 + (   +  y ) 2  ≈  x 1 +  y  ≈  y  +  xy 2  − x   y 2 + (  − x ) 2  ≈ 1   +  x   y 2 + (   +  x ) 2  ≈ 1Now, rewriting our equations of motion: m ¨ x ≈ (2 k )( − 2 x  + ( − d )(0)) − 4 k  2 x − d  x  +  xy 2  +  x  −  xy 2  + F  0  cos( ωt )= − 8 kx  +  F  0  cos( ωt )2  m ¨ y  ≈ (4 k )( − 2 y  + ( − d )(0)) − 2 k  2 y − d  y  +  xy 2  +  y  −  xy 2  + F  0  cos( ωt )= − 10 kx  +  F  0  cos( ωt )Thus we just have driven (and not damped) simple harmonic motion.The speciﬁc equations of motion we want to solve are: mx  ( t ) = − 8 kx ( t ) +  F  0  cos( ωt ) my  ( t ) = − 10 ky ( t ) +  F  0  cos( ωt )These are inhomogeneous diﬀerential equations. Morin gives the solution tothis in eqn. 3.36. The solution is: x ( t ) =  F R  cos( ωt − φ ) +  A cos( ω  t  +  θ 0 )where  ω   =   km ,  F   =  F  0 m  ,  R  =  ω  2 − ω 2 , sin φ  =  2 γω d R  →  φ  = 0, and  A  and ω 0  are set by the initial conditions. Thus, the ﬁnal solution is: x ( t ) =  A cos(   8 km t  +  θ 0 ) +  F  0 mω   2 − 8 k  cos( ωt ) y ( t ) =  A  cos(   10 km  t  +  θ  0 ) +  F  0 mω 2 − 10 k  cos( ωt )Note that these do both give the correct units of distance. If   F  0  is small,and the motion is SHM, then we note that the frequency of the motion inthe y direction is higher than in the x direction, which is to be expected,since the springs with higher constants are in the y direction. Note thatif the driving force frequency is equal to the resonance frequency, then themotion will go to inﬁnity, and  R  = 0 which is indeterminate and so ourequation does not work.2. a. We can do this by converting the cylindrical coordinates into carte-sian coordinates, and using the arc length formula for cartesian coordinates.A smooth curve with parametric equations for  x ,  y , and  z  coordinates interms of a variable  θ  that is traversed exactly once as  θ  ranges from  θ 1  to θ 2  has length    θ 2 θ 1 dθ   dxdθ 2 +  dydθ 2 +  dzdθ 2 3  In this case,  x ( θ ) =  r ( θ )cos( θ ),  y ( θ ) =  r ( θ )sin( θ ), and  z ( θ ) =  z ( r ( θ )).Thus,  x  ( θ ) =  r  cos( θ )cos( θ ) − r ( θ )sin( θ ),  y  ( θ ) =  r  ( θ )sin( θ ) +  r ( θ )cos( θ ),and  z  ( θ ) =  z  ( r ( θ )) r  ( θ ). Squaring these and plugging them in yields alength:    θ 2 θ 1 dθ   x  ( θ ) 2 +  y  ( θ ) 2 +  z  ( θ ) 2    θ 2 θ 1 dθ   ( r  cos( θ )cos( θ ) − r ( θ )sin( θ )) 2 + ( r  ( θ )sin( θ ) +  r ( θ )cos( θ )) 2 + ( z  ( r ( θ )) r  ( θ )) 2    θ 2 θ 1 dθ   r  ( θ ) 2 +  r ( θ ) 2 +  z  ( r ( θ )) 2 r  ( θ ) 2    θ 2 θ 1 dθ   r ( θ ) 2 + (1 +  z  ( r ( θ )) 2 ) r  ( θ ) 2 b. The Euler-Lagrange equation for this problem is ddθ∂L∂   ˙ r  =  ∂L∂r where  L  =   r ( θ ) 2 +  r  ( θ ) 2 +  r  ( θ ) 2 z  ( r ( θ )) 2 . Thus, we must computetwo partial derivatives and one full derivative (which we skip). ∂L∂r ( θ ) = 2 r ( θ ) + 2 r  ( θ ) 2 z  ( r ( θ )) z  ( r ( θ ))2   r ( θ ) 2 +  r  ( θ ) 2 +  r  ( θ ) 2 z  ( r ( θ )) 2 ∂L∂r  ( θ ) = 2 r  ( θ ) + 2 r  ( θ ) z  ( r ( θ )) 2 2   r ( θ ) 2 +  r  ( θ ) 2 +  r  ( θ ) 2 z  ( r ( θ )) 2 Thus, ddθ 2 r  ( θ )+2 r  ( θ ) z  ( r ( θ )) 2 2 √  r ( θ ) 2 + r  ( θ ) 2 + r  ( θ ) 2 z  ( r ( θ )) 2  =  2 r ( θ )+2 r  ( θ ) 2 z  ( r ( θ )) z  ( r ( θ ))2 √  r ( θ ) 2 + r  ( θ ) 2 + r  ( θ ) 2 z  ( r ( θ )) 2 c. We could just plug this into the above formula and obtain zero, but thealgebra is complicated enough that is simpler to start over. If   z ( r ) =  − kr ,then L  =   r ( θ ) 2 +  r  ( θ ) 2 +  k 2 r  ( θ ) 2 ∂L∂r  =  r ( θ )   r ( θ ) 2 +  r  ( θ ) 2 +  k 2 r  ( θ ) 2 4
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