Ch4 HW5 S2008

# Ch4 HW5 S2008 - Ch4 HW5 S2008 10:28 AM Web Assign Ch4 HW5...

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Unformatted text preview: Ch4 HW5 S2008 4/25/08 10:28 AM Web Assign Ch4 HW5 S2008 (Homework) Current Score: 16 out of 16 Due: Friday, February 15, 200812:00 PM EST Description Speed of sound in a solid; analytical spring- mass motion KRISTIN SEILOFF PHYS 2211 M & N Spring 08, section N05, Spring 2008 Instructor: Jennifer Curtis Instructions Reading: Sec. 4.14 - 4.16 1. 3/3 points | 3/4 submissions A ball whose mass is 1.7 kg is suspended from a spring whose stiffness is 4.5 N/m. The ball oscillates up and down with an amplitude of 10 cm. (a) What is the angular frequency ? = 1.627 1.63 radians/s (b) What is the frequency ? f = .259 0.259 hertz (cycles per second) (c) What is the period ? T = 3.862 3.86 s (d) Suppose you make it oscillate with an amplitude of 20 cm. Now what is the period ? new T = 3.862 3.86 s (e) Suppose this apparatus were taken to the Moon, where the strength of the gravitational field is only 1/6 of that on Earth. What would the period be on the Moon ? (Consider carefully how the period depends on properties of the system; look at the equation.) T = 3.862 3.86 s Solution or Explanation = ( k s /m ) 1/2 T=2 / f = 1/ T = /(2 ) The formula for the period does not contain g , so the period is not affected by gravity. http://www.webassign.net/[email protected]/[email protected] Page 1 of 4 Ch4 HW5 S2008 4/25/08 10:28 AM 2. 4/4 points | 4/4 submissions Harmonic oscillators such as spring- mass systems are the basis for accurate clocks. Let's see why. Suppose you start a vertical spring- mass system from rest with amplitude 10 cm, and it oscillates with a period of 2 seconds. Next you start this oscillator from rest with an amplitude of half the original amplitude (new amplitude = 5 cm). What happens ? The distance it goes from top to bottom (a half cycle) decreases decreases . The maximum speed decreases decreases . The period (time for a round trip) stays the same stays the same . A small amount of friction has little effect on the period of a spring- mass system, but the amplitude slowly decreases. As this oscillator slowly runs down due to friction, what happens to the "ticking" of this clock ? It ticks at the same rate It ticks at the same rate . Now consider an anharmonic oscillator. You drop a steel ball from rest onto a steel plate and it repeatedly rebounds nearly to the original height. Suppose the period of this oscillator (round- trip time) is 2 seconds. Next you start this oscillator from rest from half the original height. What happens ? The distance it goes from top to bottom (a half cycle) decreases decreases . The maximum speed decreases decreases . The period (time for a round trip) decreases decreases . As this oscillator slowly runs down due to friction, what happens to the "ticking" of this clock ? It ticks faster It ticks faster . This is why harmonic oscillators are so important in a wide range of scientific and engineering applications. 3. 3/3 points | 2/4 submissions A mass of 2.2 kg is connected to a horizontal spring whose stiffness is 6.5 N/m. When the spring is relaxed, x = 0. The spring is stretched so that the initial value of x = + 0.16 m. The mass is released from rest at time t = 0. Remember that when the argument of a trigonometric function is in radians, on a calculator you have to switch the calculator to radians or convert the radians to degrees. Predict the position x when t = 1.28 s: x t = 1.28 s = - .094 http://www.webassign.net/[email protected]/[email protected] Page 2 of 4 Ch4 HW5 S2008 4/25/08 10:28 AM - 0.0942 m Solution or Explanation x = A cos( t ) = ( k s / m ) 1/2 Since there is no initial velocity, A = x 0 4. 4/4 points | 1/4 submissions In Problem 4.P.76 (page 157) we found the effective spring stiffness corresponding to the interatomic force for aluminum and lead. Let's assume for the moment that, very roughly, other atoms have similar values. (a) What is the (very) approximate frequency f for the oscillation ("vibration") of H2 , a hydrogen molecule containing two hydrogen atoms? Remember that frequency is defined as the number of complete cycles per second or "hertz": f = 1/ T. There is no one correct answer, since we're just trying to calculate the frequency approximately. However, just because we're looking for an approximate result doesn't mean that all answers are correct! Calculations that are wildly in disagreement with what physics would predict for this situation will be counted wrong. f = 1.41e13 1.4e13 cycles/s (hertz) (b) What is the (very) approximate frequency f for the vibration of O 2 , an oxygen molecule containing two oxygen atoms? f = 2e12 3.5e12 cycles/s (hertz) (c) What is the approximate vibration frequency f of D2 , a molecule both of whose atoms are deuterium atoms (that is, each nucleus has one proton and one neutron)? f = 1e13 1e13 cycles/s (hertz) (d) Which of the following statements are true? (Select all that apply.) [x] [x] Neither of the estimated frequencies for D2 and H2 is accurate, but the ratio of the D2 frequency to the H2 frequency is quite accurate, because the "spring" represents the interatomic force, which is nearly the same for atoms with similar chemical structure (number of electrons). [_] [_] The estimated frequencies for D2 and H2 are both quite accurate, because these are simple molecules, and the effective spring stiffness is expected to be the same as we found inside a block of metal. [_] [_] The estimated frequency for O 2 is quite accurate because the mass of an oxygen atom is similar to the mass of an aluminum atom. [x] [x] The true vibration frequency for D2 is lower than the true vibration frequency for H2 , because the mass is larger but the effective "spring" stiffness is nearly the same, since it is related to the electronic structure, which is nearly the same for D2 and H2 . Solution or Explanation Here is a complete solution. http://www.webassign.net/[email protected]/[email protected] Page 3 of 4 Ch4 HW5 S2008 4/25/08 10:28 AM 5. 2/2 points | 4/4 submissions Uranium - 238 (U 238 ) has three more neutrons in its nucleus than uranium- 235 (U 235 ). Chemically, the atoms of these two isotopes behave in essentially identical ways, since the clouds of 92 electrons are nearly identical. Which of the following statements are true, and are needed to fully support the conclusion about the speed of sound? More than one supporting statement may be needed. [_] [_] The speed of sound in U238 is higher than the speed of sound in U235 . [x] [x] The effective stiffness of the interatomic "spring" is the same in the two isotopes, since this "spring" is a model for the interactions of the outer electrons, which are the same for different isotopes of the same atom (only the nuclei are different). [_] [_] The speed of sound is lower in U238 because an atom with more mass in the nucleus is bigger, so the distance between neighboring atoms is larger. [x] [x] The speed of sound in U238 is lower than the speed of sound in U235 . [_] [_] The speed of sound in U238 is the same as the speed of sound in U235 . [_] [_] The atomic mass is larger in U238 , so the speed is higher, because the atoms can push harder on each other if they have more mass. [_] [_] The reason why the speed of sound is the same in the two isotopes is that the speed of sound just depends on the interatomic force, which is the same for atoms that differ only in their nuclei. [x] [x] The atomic mass is larger in U238 , and the speed of sound is inversely proportional to the square root of the mass, so the speed is lower if the effective stiffness of the interatomic "spring" is the same. http://www.webassign.net/[email protected]/[email protected] Page 4 of 4 ...
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