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Questions 1, 2, and 3 are short free-response questions that require about 13 minutes to answer and are worth 8 points. Questions 4 and 5 are long free-response questions that require about 25 minutes each to answer and are worth 13 points each. Show your work for each part in the space provided after that part.(DIAGRAM IS NOT AVAILABLE TO COPY)A horizontal spring with a force constant of 40 $\mathrm{N} / \mathrm{m}$ is attached a 0.1 $\mathrm{kg}$ block.(a) If the block is pulled to a distance of 0.5 m and released, what is the maximum speed of the block?(b) What is the frequency of the oscillations?(c) If the spring were flipped vertically and attached to the ground with the block placed on top, how would the natural length of the spring change?(d) How does the frequency of the oscillations of the vertical spring block oscillator compare with that when it was placed horizontally?

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Chapter 14

Practice Test 4

Section 2

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Directions: Questions 1, 2…

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A block of mass M oscillat…

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A horizontal spring with a…

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You have a spring that str…

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A horizontal spring attach…

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Review problem. A block ha…

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Vibrations of a spring Sup…

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A horizontal block-spring …

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(III) $\mathrm{A} 2.0$ -kg…

here we are considering a simple spring past system to look at some of the quantities that are involved when a mass starts oscillating on a spring. So we're going to start off with a mass oscillating horizontally. We are told its mass like one kg and it spring constant for the spring is 40 newtons per meter. And what I suggest is we immediately know the frequency with which this mass will oscillate. We'll pretend it's on a frictionless surface so there is no friction. Um But the angular frequency omega is simply equal to the square root of K over em. So those two parameters govern how quickly it oscillates if we want that in terms of frequency in cycles per second. There's a simple relationship. Um And if we use the numbers given 40.1, What we get basically is 20 radiance per second for the angular frequency. And that means are uh frequency in cycles per second Is just that divided by two pi. So that would be 10 over by cycles per second. So roughly three cycles per second. So we know already how this thing will oscillate wants to get set into motion but we can figure out another parameter. Let's assume that we stretch the block immediately to the right by an amount of five centimeters. It's a reasonable amount to stretch that spring. But we can now calculate is not only the frequency of oscillation, but how fast it is going to oscillate back and forth. We can simply use conservation of energy that the original elastic energy that's in that system. When you stretch it gets transferred all into kinetic energy when that mass goes through the equilibrium position. So we'll consider that the point of highest velocity through the equilibrium position where the spring is neither stretched nor compressed. Um And of course kinetic energy is 1/2. V. Squared. So we could figure out the velocity of this mess. Yeah. Mhm. Some things will cancel their nicely and the velocity final will be equal to the square root of K over em. Um And then the amount of stretch so notice that that final velocity is going to be related to the omega, the angular frequency and that's kind of interesting but you need a stretch in order to turn this into a velocity. Um And we knew that that Initial Omega was 20 radiance per second And the s. is five cm Or 5.5 m. So that will translate into a final velocity Of 1.0 m/s. Okay, finally we can ask what if instead of um oscillating the mass horizontally, what if we put it on top of the spring? So we'll assume there's some sort of little platform of something that we can put the mass on to load the spring down vertically. The thing to note is that you really haven't changed the stiffness or the mass. Hopefully compressing it will not um do anything to the spring constant. Um Ideal Spring would not work that way. So the Omega is still 20 rads per second. And about all that has happened is you've changed the equilibrium position. Hook's law tells us that okay times the amount of compression, so just sitting on their the mass is going to compress the spring. But according to Hook's law, that compression will change the length of the spring KS equals M. J. In equilibrium. And if we saw for S it's roughly 0.1 times 10 over 40 meters, 4.025 m. So the result of putting the mass on the top is to shorten the spring bye. A couple of centimeters. And hopefully that will not change the spring constant, not for an ideal spring.

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Directions: Questions 1, 2, and 3 are short free-response questions that req…

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