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Unformatted text preview: 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 = ⇒ 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 ) = Aω sin ( ωt ) ⇒ v ( t ) = Aω sin ( ωt ) ⇒ a ( t ) = d dt v ( t ) = Aω 2 cos ( ωt ) ⇒ a ( t ) = Aω 2 cos ( ωt ) . 4. Plots of x , v , and a are shown below: 2 3 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 )] ⇒ Aω 2 cos ( ωt ) Aω 2 cos ( ωt ) = 0. X 6. ω = q k m ⇒ [ ω ] = q [ k ] m = q [ F ] mL = q m [ a ] mL = q L LT 2 = q 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 r k m . T = 1 f ⇒ T = 2 π r m k . 9. T = 2 π p m k → T = 2 π q 4 m k = 4 π p m k = 2 T ⇒ T increases by a factor of 2. f = 1 2 π q k m → f = 1 2 π q k 4 m = 1 4 π q k m = 1 2 f ⇒ f decreases by a factor of 2. ω = q k m → ω = q k 4 m = 1 2 q k m = 1 2 ω ⇒ ω decreases by a factor of 2. 10. T = 2 π p m k → T = 2 π p m 4 k = π p m k = 1 2 T ⇒ T decreases by a factor of 2. f = 1 2 π q 4 k m → f = 1 2 π q 4 k m = 1 π q k m = 2 f ⇒ f increases by a factor of 2. ω = q k m → ω = q 4 k m = 2 q 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 = 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. We may derive the expression for potential energy from a force F ( x ) using the 4 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 ) = R 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...
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This note was uploaded on 04/12/2011 for the course PHYSICS 6b taught by Professor Gruner during the Spring '10 term at UCLA.
 Spring '10
 GRUNER
 Physics

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