Simple harmonic motion-solutions

Simple harmonic motion-solutions - le (txl65) – Simple...

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Unformatted text preview: le (txl65) – Simple harmonic motion – mohanty – (81440) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points The mass of the deuterium molecule D 2 is twice that of the hydrogen molecule H 2 . If the vibrational frequency of H 2 is 1 . 22 × 10 14 Hz, what is the vibrational frequency of D 2 , assuming that the “spring constant” of attracting forces is the same for the two species? Correct answer: 8 . 6267 × 10 13 Hz. Explanation: Let : M D = 2 M H . The angular frequencies depend only on spring constant and mass: ω = radicalbigg k M ∝ radicalbigg 1 M ω D = radicalBigg k M D ω H = radicalBigg k M H The spring constants k are the same, so ω D ω H = radicalBigg M H M D = radicalBigg M H 2 M H = radicalbigg 1 2 = 1 √ 2 . The linear frequency is f = ω 2 π ∝ ω, so f D f H = ω D ω H = 1 √ 2 f D = f H √ 2 = 1 . 22 × 10 14 Hz √ 2 = 8 . 6267 × 10 13 Hz . 002 10.0 points An 81 . 6 g mass is attached to a horizontal spring with a spring constant of 12 . 2 N / m and released from rest with an amplitude of 28 . 1 cm. What is the velocity of the mass when it is halfway to the equilibrium position if the surface is frictionless? Correct answer: 2 . 97558 m / s. Explanation: Let : m = 81 . 6 g , k = 12 . 2 N / m , and A = 28 . 1 cm . x = A 2 = 0 . 1405 m v = radicalbigg k m ( A 2- x 2 ) = radicalBigg 12 . 2 N / m . 0816 kg [(0 . 281 m) 2- (0 . 1405 m) 2 ] = 2 . 97558 m / s . 003 10.0 points An automobile having a mass of 820 kg is driven into a brick wall in a safety test. The bumper behaves like a spring of con- stant 7 . 5 × 10 6 N / m and compresses 1 . 4 cm as the car is brought to rest. What was the speed of the car before im- pact, assuming no energy was lost during im- pact with the wall? Correct answer: 1 . 33891 m / s. Explanation: Let : m = 820 kg , k = 7 . 5 × 10 6 N / m , and x = 1 . 4 cm . The potential energy of spring is U spring = 1 2 kx 2 . le (txl65) – Simple harmonic motion – mohanty – (81440) 2 Initially the car (mass m ) has a speed v and therefore a kinetic energy of K i = 1 2 mv 2 ; after the collision, the car is at rest but the bumper, approximated by a spring, has a po- tential energy U spring . By conservation of energy, K i + U i = K f + U f 1 2 mv 2 = 1 2 kx 2 v = radicalbigg k m x =...
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This note was uploaded on 04/05/2011 for the course SD ds taught by Professor Sdsd during the Spring '11 term at Tarrant County.

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Simple harmonic motion-solutions - le (txl65) – Simple...

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