Physics 260 Homework Assignment 2 1 PSE6 15 CQ 004 a In simple harmonic motion ont half of the time the velocity is in the same direction as the displacement away from equilibrium b Velocity and acceleration are in the same direction half of the time c Acceleration is always opposite to the position vector and never in the same direction 2 PSE6 15 CQ 010 If the pendulum s length is doubled then Lf 2Li From the equation of period of a simple pendulum we have 2 Lf 2Li Li Tf 2 2 2 2 2Ti f g g g s s s If the mass of the suspended bob is doubled the period of the pendulum is unaffected because the period equation doesn t depends on the mass of the suspended bob 3 PSE6 15 CQ 011 a the elevator accelerates upward Tension provided by the string need to overcome the gravity and provide enough force for the mass to accelerate upward at acceleration a so gef f g a From the period equation we know that q T 2 L gef f Since gef f g T decreases b the elevator accelerates downward gef f g a therefore T increases c the elevator moves with constant velocity there is no net acceleration gef f stays same So does the period 1 4 PSE6 15 P 002 a Plug in t 0 and calculate x b v dx 2 5 00cm sin 2t dt 6 Plug in t 0 and calculate t c a dv 4 5 00cm cos 2t dt 6 Plug in t 0 and calculate a d The amplitude is the number in front of centimeter The value before t gives the angular frequency of the motion T 2 5 PSE6 14 P 007 a uses the number of seconds divide by the number of complete cycles to get period b There are two ways to find this We can either use one over period to get the frequency or use the number of complete cycles divide by the number of seconds c 6 2 f 2 T PSE6 14 P 009 2 T k 7 s k m 4m 2 T2 PSE6 14 P 015 a Energy is conserved for the block spring system between the maximum displacement and the half maximum points K U max K U half max 2 AT half maximum xhalf max 12 A 1 1 1 1 1 0 kA2 mv 2 kx2half max mv 2 kA2 2 2 2 2 8 Rearrange the terms and solve for v s v 8 3kA2 4m PSE6 14 P 018 a E K U 0 21 kA2 The amplitude needs to be converted to meters 2 E 12 mvmax b vmax q 2E m c The block has maximum acceleration when it is at maximum displacement mamax F kA 9 amax kA m PSE6 14 P 032 a The string tension must support the weight of the bob accelerate it upward and also provide the restoring force just as if the elevator were at rest in a gravity field gef f g a same argument in problem 3 s T 2 s L gef f 2 L g a b The concept used here is exactly the same as part a except gef f g a c The effective g should now be gef f use the period equation used in part a 10 g 2 a2 vector addition Again PSE6 14 P 045 a We are given the angular frequency of the motion 2 So T 2 q b 0 k m We are given F0 3 00N and b 0 Applying Equation 15 36 in the textbook we have the amplitude of a driven oscillator as follows F0 m A q 2 02 2 b 2 m 3 11 PSE6 14 P 063 q L g a T 2 b E 21 mvi2 c Energy is conserved Initially the pendulum has only kinetic energy y 0 At maximum angular displacement all the kinetic energy transform into potential energy y h Hence 1 mgh mvi2 2 h vi2 2g Also L L cos h So cos 1 h L cos 1 1 4 h L
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