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When a mass is attached to a vertical spring, the spring is stretched a distance d. The mass is then pulled do?

When a mass is attached to a vertical spring, the spring is stretched a distance d. The mass is then pulled do? Topic: Case springs
June 20, 2019 / By Kacey
Question: i really need to know the steps and equations used to get the answer When a mass is attached to a vertical spring, the spring is stretched a distance d. The mass is then pulled down from this position and released. It undergoes 54 oscillations in 30.2 s. What was the distance d?
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Best Answers: When a mass is attached to a vertical spring, the spring is stretched a distance d. The mass is then pulled do?

Gretta Gretta | 1 day ago
OK 1)A spring has a length =L with no mass and K=new/m it's an unstretched spring 2)You add a mass "m" attached at the end (getting a +Xst,then it's on an static equilibrium) This is the position of the equilibrium line... Xo-Xo called also equilibrium poisition 3)The forces acting on the spring are: w=mg and the reaction (Fo) of it (means the force excerted by the spring) 4)so we got: W=Fo....W=mg and Fo= K*Xst hence ....mg=K*Xst........level.......xo-xo 5) After this if you displace the mass plus through a distance X1 from it's equilibrium position and release it, with no velocity will get a vibration of amplitud "A" so,the mass will move back and forth through its equilibrium position; in this case the new forcé excerted by the spring is (F1) F1= K*X1......level x1-x1 6) Energy Eu= 1/2*K*(X1)² ........Ek=1/m*v1² At xo-xo position..... vo=max....Eu=o........A=0 At the border.. v=0 Eu=max...x=A well, these are the basic knowledges you have to know in this issue Now,hitting the problem F=kd.........mg=Kd....g/d=K/m.....(1) At equilibrium position.x=0.......xo-xo Ek=0.5*m*Vmax²......(2)...Ep=0...Vmax=0 and we know....................Vmax=w*A At extreme side x=A..... amplitud Ep=0.5 K A².............(3).....EK=0. doing...(2)=(3) 0.5 m*ω²A²=0.5 K A².....mω²A²=K A²..ω.²=K/m. taking from (1)......g/d=K/m ω²=g/d,....but.... ω= 54 oscil*π/30.2sec=1.788π*rad/s 3.197*π²=g/d.......d=9.81/3.197*π²=3.07...‡ Luck hope helps Cuglier
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Gretta Originally Answered: When a mass is attached to a vertical spring, the spring is stretched a distance d. The mass is then pulled do?
OK 1)A spring has a length =L with no mass and K=new/m it's an unstretched spring 2)You add a mass "m" attached at the end (getting a +Xst,then it's on an static equilibrium) This is the position of the equilibrium line... Xo-Xo called also equilibrium poisition 3)The forces acting on the spring are: w=mg and the reaction (Fo) of it (means the force excerted by the spring) 4)so we got: W=Fo....W=mg and Fo= K*Xst hence ....mg=K*Xst........level.......xo-xo 5) After this if you displace the mass plus through a distance X1 from it's equilibrium position and release it, with no velocity will get a vibration of amplitud "A" so,the mass will move back and forth through its equilibrium position; in this case the new forcé excerted by the spring is (F1) F1= K*X1......level x1-x1 6) Energy Eu= 1/2*K*(X1)² ........Ek=1/m*v1² At xo-xo position..... vo=max....Eu=o........A=0 At the border.. v=0 Eu=max...x=A well, these are the basic knowledges you have to know in this issue Now,hitting the problem F=kd.........mg=Kd....g/d=K/m.....(1) At equilibrium position.x=0.......xo-xo Ek=0.5*m*Vmax²......(2)...Ep=0...Vmax=0 and we know....................Vmax=w*A At extreme side x=A..... amplitud Ep=0.5 K A².............(3).....EK=0. doing...(2)=(3) 0.5 m*ω²A²=0.5 K A².....mω²A²=K A²..ω.²=K/m. taking from (1)......g/d=K/m ω²=g/d,....but.... ω= 54 oscil*π/30.2sec=1.788π*rad/s 3.197*π²=g/d.......d=9.81/3.197*π²=3.07...‡ Luck hope helps Cuglier
Gretta Originally Answered: When a mass is attached to a vertical spring, the spring is stretched a distance d. The mass is then pulled do?
The most important thing to wrap your head around is this. The force of a spring equals the force of gravity when the spring is pulled to the maximum distance from equilibrium (d). (Imagine the force of gravity pulling the mass down when the spring is compressed, so the spring pushes down at that same time.) Equations: Force equals mass times gravity. F=mg Force of spring equals negative spring constant times distance from equilibrium. F'=-kx At the maximum distance from equilibrium x is equal to d, so we are trying to solve for x. The key here is setting these equal. F'=-F therefore mg=kx Now I know that mass is hanging out and being annoying, so to get rid of it we have to replace k with something else. Remember frequency equals 2*pi*angular frequency? (f=2*pi*w) where f is frequency in hertz and w (omega) is in radians per second. Angular frequency (w or omega) equals the square root of (the spring constant divided by the mass). Here is where we find mass and the spring constant for our mg=kx equation. This can be re-written and substituted for k. m*(w^2)=k Finally, the mass cancels out and we can solve for x. Here's the magic equation. d=g/(w^2) So, first find your frequency (f) which is just the oscilations divided by the seconds. Then multiply f by 2*pi which will give you w (omega). Now divide gravity (g=9.82) by w squared to get the distance in meters. (54/30.2)=f 2*pi*f=w g/(w^2)=d g=9.81 9.81/((2*pi*(54/30.2))^2) = 0.0777 So the final answer for your example is 0.0777 meters.

Diamond Diamond
The most important thing to wrap your head around is this. The force of a spring equals the force of gravity when the spring is pulled to the maximum distance from equilibrium (d). (Imagine the force of gravity pulling the mass down when the spring is compressed, so the spring pushes down at that same time.) Equations: Force equals mass times gravity. F=mg Force of spring equals negative spring constant times distance from equilibrium. F'=-kx At the maximum distance from equilibrium x is equal to d, so we are trying to solve for x. The key here is setting these equal. F'=-F therefore mg=kx Now I know that mass is hanging out and being annoying, so to get rid of it we have to replace k with something else. Remember frequency equals 2*pi*angular frequency? (f=2*pi*w) where f is frequency in hertz and w (omega) is in radians per second. Angular frequency (w or omega) equals the square root of (the spring constant divided by the mass). Here is where we find mass and the spring constant for our mg=kx equation. This can be re-written and substituted for k. m*(w^2)=k Finally, the mass cancels out and we can solve for x. Here's the magic equation. d=g/(w^2) So, first find your frequency (f) which is just the oscilations divided by the seconds. Then multiply f by 2*pi which will give you w (omega). Now divide gravity (g=9.82) by w squared to get the distance in meters. (54/30.2)=f 2*pi*f=w g/(w^2)=d g=9.81 9.81/((2*pi*(54/30.2))^2) = 0.0777 So the final answer for your example is 0.0777 meters.
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Diamond Originally Answered: How does the amplitude of a spring system depend on mass?
(Ignoring all air and spring friction) Amplitude is the magnitude of change in position. In simple harmonic motion, such as that of a weight hanging from a spring, the mass of the object has no influence on the amplitude. If you pull the weight 1 inch down from equilibrium, it will travel back to equilibrium and then 1 inch above equilibrium then back to equilibrium and 1 inch below equilibrium. . .on and on. This is because of conservation of energy. When you pull down on the weight, you are loading the system with potential energy (from the spring). As the weight crosses the equilibrium point, the force of gravity and spring force are equal and opposite so the potential energy is zero but all of that energy still exists in the form of kinetic energy (motion of the weight traveling up). When the weight reaches 1 inch above equilibrium, it has expended all of it's kinetic energy but it now has a height above equilibrium so it now has all of that energy in the form of potential energy (from gravity). You can find the amplitude of this setup without knowing the mass of the weight at all. Some possible reasons for your lab observation: The above explanation relies on an ideal spring and no friction. A poor spring may lose energy faster with a heavy weight. In the real world, some of that energy is lost to air and spring friction with every pass. A heavier weight may help it to stay in motion longer if it has a good spring. This is because the forces of friction remain essentially the same while the force mg is bigger.
Diamond Originally Answered: How does the amplitude of a spring system depend on mass?
Your human hand is inaccuarate. You don't have a Newton-gauge or a Joule-gauge attached to any wrist watch you can use. At most you can position the initial conditions...and even then it is inaccurate. There shouldn't necessarily be any trend between mass and amplitude unless you are mixing possible amplitude-types. Hint: maximum energy/maximum restoring force/maximum displacement from equilibrium/maximum speed are all valid amplitude types to consider.

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