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Unformatted text preview: Sample Exam 1 Solutions Antoine Calvez, Eric Angle Physics Department, University of California Los Angeles. * (Dated: October 17, 2006) * Electronic address: acalvez@ucla.edu,angle@physics.ucla.edu 1 I. SAMPLE EXAM PROBLEM 1. 1. Hookes Law is F x = kx . Newtons 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 = d 2 x dt 2 + k m x = 0, or d 2 x dt 2 + 2 x = 0 , where 2 = k m . 3. x ( t ) = A cos ( t ) v ( t ) = d dt x ( t ) = A sin ( t ) v ( t ) = A sin ( t ) a ( t ) = d dt v ( t ) = A 2 cos ( t ) a ( t ) = A 2 cos ( t ) . 4. Plots of x , v , and a are shown below: 2 3 5. We aim to show that if x = A cos ( t ), d 2 x dt 2 + 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 ( t )] + 2 [ A cos ( t )] A 2 cos ( t ) A 2 cos ( t ) = 0. X 6. = q k m [ ] = q [ k ] m = q [ F ] mL = q m [ a ] mL = q L LT 2 = q 1 T 2 = 1 T [ ] = 1 T . Angular frequency, , should have units of 1 T , since this is the same dimension of frequency f , and we know and f have the same dimension from the expression = 2 f . 7. f = 1 T = 2 . T , the period, is the time one complete oscillation takes. f , the frequency, is the number of complete oscillations that occur per second. , 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 = 2 f = 1 2 r k m . T = 1 f T = 2 r m k . 9. T = 2 p m k T = 2 q 4 m k = 4 p m k = 2 T T increases by a factor of 2. f = 1 2 q k m f = 1 2 q k 4 m = 1 4 q k m = 1 2 f f decreases by a factor of 2. = q k m = q k 4 m = 1 2 q k m = 1 2 decreases by a factor of 2. 10. T = 2 p m k T = 2 p m 4 k = p m k = 1 2 T T decreases by a factor of 2. f = 1 2 q 4 k m f = 1 2 q 4 k m = 1 q k m = 2 f f increases by a factor of 2. = q k m = q 4 k m = 2 q k m = 2 increases by a factor of 2. 11. None of the physical quantities T , f , or 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 = R x 2 x 1 dxF ( x ). Since F ( x ) = m dv dt , W = m R x 2 x 1 dx dv dt = W = m R 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 workenergy theorem....
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This note was uploaded on 04/29/2008 for the course PHYS 6B taught by Professor Wu during the Winter '08 term at UCLA.
 Winter '08
 Wu
 Physics

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