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mt1_sample_solution

mt1_sample_solution - Sample Exam 1 Solutions Antoine...

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Sample Exam 1 Solutions Antoine Calvez, Eric Angle Physics Department, University of California Los Angeles. * (Dated: October 17, 2006) * Electronic address: [email protected],[email protected] 1
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I. SAMPLE EXAM PROBLEM 1. 1. Hooke’s Law is F x = - kx . Newton’s Second Law is F x = m d 2 x dt 2 . [1] 2. Setting these expressions equal to one another gives F x = - kx = m d 2 x dt 2 m d 2 x dt 2 + kx = 0 d 2 x dt 2 + k m x = 0, or d 2 x dt 2 + ω 2 x = 0 , where ω 2 = k m . 3. x ( t ) = - A cos ( ωt ) v ( t ) = d dt x ( t ) = sin ( ωt ) v ( t ) = sin ( ωt ) a ( t ) = d dt v ( t ) = 2 cos ( ωt ) a ( t ) = 2 cos ( ωt ) . 4. Plots of x , v , and a are shown below: 2
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5. We aim to show that if x = - A cos ( ωt ), d 2 x dt 2 + ω 2 x = 0 is satisfied. Plug- ging this expression for x into the left hand side of the differential equation gives d 2 dt 2 [ - A cos ( ωt )] + ω 2 [ - A cos ( ωt )] 2 cos ( ωt ) - 2 cos ( ωt ) = 0. 6. ω = k m [ ω ] = [ k ] m = [ F ] mL = m [ a ] mL = L LT 2 = 1 T 2 = 1 T [ ω ] = 1 T . Angular frequency, ω , should have units of 1 T , since this is the same dimension of frequency f , and we know ω and f have the same dimension from the expression ω = 2 πf . 7. f = 1 T = ω 2 π . T , the period, is the time one complete oscillation takes. f , the frequency, is the number of complete oscillations that occur per second. ω , the angular frequency, is the number of radians traversed per second by the argument of the sinusoidal function used in the position function of the oscillator. 8. f = ω 2 π f = 1 2 π k m . T = 1 f T = 2 π m k . 9. T = 2 π m k T = 2 π 4 m k = 4 π m k = 2 T T increases by a factor of 2. f = 1 2 π k m f = 1 2 π k 4 m = 1 4 π k m = 1 2 f f decreases by a factor of 2. ω = k m ω = k 4 m = 1 2 k m = 1 2 ω ω decreases by a factor of 2. 10. T = 2 π m k T = 2 π m 4 k = π m k = 1 2 T T decreases by a factor of 2. f = 1 2 π 4 k m f = 1 2 π 4 k m = 1 π k m = 2 f f increases by a factor of 2. ω = k m ω = 4 k m = 2 k m = 2 ω ω increases by a factor of 2. 11. None of the physical quantities T , f , or ω depend on A , so if the amplitude changes, none of these physical quantities change. 12. K = 1 2 m dx dt 2 , U = 1 2 kx 2 . We can derive the expression for kinetic energy by first noting that the work done on a particle moving from x = x 1 to x = x 2 subject to a force F ( x ) is W = x 2 x 1 dxF ( x ). Since F ( x ) = m dv dt , W = m x 2 x 1 dx dv dt = W = m v 2 v 1 vdv = 1 2 m ( v 2 2 - v 1 2 ). If we define K = 1 2 mv 2 , W = K 2 - K 2 , which is the work-energy theorem. We may derive the expression for potential energy from a force F ( x ) using the 4
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15. relation F ( x ) = - dU ( x ) dx . (Note: only special forces may always be written as the derivative of a potential function, and they are conservative forces.) Using this expression, dU ( x ) = - F ( x ) dx . Hooke’s Law gives F ( x ) = - kx , so in this case U ( x ) = kxdx = 1 2 kx 2 + U (0). We may choose U (0) to be anything, and it is typically set to be zero, so that U ( x ) = 1 2 kx 2 .
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