HW10a - Version One Homework 10 savrasov 39822 Dec 05, 2007...

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Version One – Homework 10 – savrasov – 39822 – Dec 05, 2007 1 This print-out should have 9 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. The due time is Central time. Tipler PSE5 14 104 13:01, calculus, numeric, > 1 min, normal. 001 (part 1 oF 1) 10 points A wooden cube with edge length 5 cm and mass 62 . 5 g ±oats in water with one oF its Faces parallel to the water surFace. ²ind the period oF oscillation in the vertical direction iF the cube is pushed down slightly. The density oF the water is 1 g / cc, and the acceleration oF gravity is 9 . 81 m / s 2 . Correct answer: 0 . 317187 s. Explanation: Let : m = 62 . 5 g , a = 5 cm , ρ = 1 g / cc , and g = 9 . 81 m / s 2 . The density oF the wood is ρ wood = m a 3 = 62 . 5 g (5 cm) 3 = 0 . 5 g / cc . Applying X F y = 0 to the cube when it is ±oating in the water, m g - F B = 0 . Applying X T y = m a y to the cube when it is pushed down a small distance y , m g - F 0 B = m a y . Subtracting , F B - F 0 B = m a y = Δ F B . ²or y ¿ 1 , Δ F B = dF B = - ρ V g = - a 2 ρ g y = m d 2 y dt 2 . The equation oF motion is m d 2 y dt 2 = - a 2 ρ g y d 2 y dt 2 = - a 2 ρ g m y = - ω 2 y ω = a r ρ g m , so the period oF the oscillation is T = 2 π ω = 2 π a r m ρ g = 2 π 5 cm s 62 . 5 g (1 g / cc)(9 . 81 m / s 2 ) × r 1m 100cm = 0 . 317187 s . keywords: Vibrating D2 Molecule 13:02, trigonometry, numeric, > 1 min, nor- mal. 002 (part 1 oF 1) 10 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 . 3 × 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: 9 . 19239 × 10 13 Hz. Explanation: Let : M D = 2 M H . The angular Frequencies depend only on spring constant and mass: ω = r k M r 1 M ω D = s k M D
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Version One – Homework 10 – savrasov – 39822 – Dec 05, 2007 2 ω H = s k M H The spring constants k are the same, so ω D ω H = s M H M D = s M H 2 M H = r 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 . 3 × 10 14 Hz 2 = 9 . 19239 × 10 13 Hz . keywords: Uniform Rod as a Pendulum 05 13:06, calculus, numeric, > 1 min, normal. 003
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HW10a - Version One Homework 10 savrasov 39822 Dec 05, 2007...

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