In-class problems 1 Solutions

# In-class problems 1 Solutions - (b The maximum speed of the...

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Name Student ID Section In-Class Problem, Week 2 Solution 8 April 2008 Total Points Possible: 20 1. The position of simple harmonic oscillator varies periodically in time according to the relation x ( t ) = A cos( ωt + φ ). Find the following quantities, leaving your answers in terms of the amplitude A and the frequency ω . (a) The maximum speed v ( t ) (2.5 points) v = dx dt = - ωA sin( ωt ) (+ 1 pt) max sin( ωt ) = ± 1 (+ 0.5 pts) v max = ωA (+ 1 pt) (b) The maximum acceleration a ( t ) (2.5 points) a = dx 2 dt 2 = dv dt = - ω 2 A cos( ωt ) (+ 1 pt) max cos( ωt ) = ± 1 (+ 0.5 pts) a max = ω 2 A (+ 1 pt) 2. A block-spring system oscillates with an amplitude of 2.25 cm. The spring constant is 283 N/m, and the mass of the block is 0.435 kg. Determine (a) The mechanical energy of the system (5 points) E = 1 2 kA 2 = 1 2 (283 N/m )(0 . 0225 m ) 2 (+ 3 pts) = 0 . 0716 J (numerical answer = + 1 pt, correct units = + 1 pt)
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Unformatted text preview: (b) The maximum speed of the block (5 points) v max = ωA = q k m A = q 283 N/m . 435 kg (0 . 0225 m ) (+ 3 pts) = 0 . 574 m/s (numerical answer = + 1 pt, correct units = + 1 pt) (c) The maximum acceleration of the block (5 points) a max = ω 2 A = k m A = 283 N/m . 435 kg (0 . 0225 m ) (+ 3 pts) = 14 . 6 m/s 2 (numerical answer = + 1 pt, correct units = + 1 pt) Some possibly useful equations: d cos( ωt ) dt =-ω sin( ωt ) d sin( ωt ) dt = ω cos( ωt ) ω = 2 πf = q k m T = 1 f = 2 π ω E total = K + U = 1 2 kA 2 K = 1 2 mv 2 U = 1 2 kx 2 v ( t ) = dx ( t ) dt a ( t ) = d 2 x ( t ) dt 2 = dv ( t ) dt F =-kx ω 2 = g L 1...
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