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Lab Report

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LAB REPORT 4
Hooke’s Law Objectives * To perform an experiment to measure the spring constant of a coil spring.

Theory * Hooke's Law states that if the distortion of an elastic body is not too large, the force tending to restore the body to equilibrium is proportional to the displacement of the body from equilibrium. F = - k r where F is a restoring force, k is a constant of proportionality and x is the distance the object has been displaced from its equilibrium position. The minus sign signifies that the restoring force acts in the opposite direction to the displacement of the body from the equilibrium position. So, we can translate this formula into: F= KsX * If a body, which obeys Hooke’s Law, is displaced from equilibrium and released, the body will undergo “simple harmonic motion”. From the experiment, we can get the T that is one oscillation of the spring, so we can get KT by: KT=4π2Mt/T2

Procedures

Way One * Set up the apparatus as illustrated. * Place 200 grams on the spring and record the position of the bottom of the weight pan along the meter stick. * Repeat the step above by using 400 grams, 600 grams, 800 grams and 1000 grams, record the positions of all these masses. * Computer the force on the spring F by using the formula F= mg * Plot a graph of force, F VS Position, x using the values of F and x in the data chart and compute the slope the graph * Pick up any two points of the line, that is (x1 y1), (x2 y2). Computer the slope of graph: Slope = (y2- y1)/( x2- x1) that is also the spring constant KS in Hook`s law by using formula F=KSX

Way Two * Place about 800 grams of mass on the spring and start the spring oscillating * With a stopwatch, record the time, t, that it takes for the spring to oscillate 20 times. * Compute the period, T for one oscillation of the spring using
T=t/20
* Computer the spring constant of the spring, KT using the formula, where Mt is 800 grams KT=4π2Mt/T2
% diff * use the records KT and KS, get the % diff by formula % diff = (KS – KT)*100/[( KS +KT)/2]

Data
Part One Mass m(grams) | Position of spring x(cm) | Force on spring F(dynes) | 200 | 25 | 196000 | 400 | 33.5 | 392000 | 600 | 41.5 | 588000 | 800 | 49.5 | 784000 | 1000 | 57 | 980000 |

* Pick two points on the line of the graph (50, 800000) and (40, 550000) then get * Slope of graph= Ks=25000

Part Two * t (it takes for the spring to oscillate 20 times)=22.66 * T (it takes for the spring to oscillate 1 time) =1.133 * KT=24603

Part Three * % diff = 1.6

Calculations
Part One * Force on spring F(dynes)
200 grams: 200 g*980cm/s2=196000 dynes
400 grams: 400 g*980cm/s2=392000 dynes
600 grams: 600 g*980cm/s2=588000 dynes
800 grams: 800 g*980cm/s2=784000 dynes
1000 grams: 1000 g*980cm/s2=980000 dynes * Pick two points on the line of the graph (50, 800000) and (40, 550000), so
Slope of graph=KS=(800000-550000)/(50-40)=25000

Part Two * For t= 22.66, so T= 22.66/20=1.133 * We have already get MT=800 g, so KT=4π2Mt/T2=4π2*800/1.1332=24603

Part Three
% diff = (KS – KT)*100/[( KS + KT)/2]=(25000-24603)*100/[(25000+24603)/2]=1.6% Error Analysis and improvements * The % diff is 1.6%, it might be the two points we pick up are not so close to the line, that makes the result of the slope, also the result of the KS is not correctly. * The other reason can cause the % diff is the time we get is not exact, that makes the KT is little far from the perfect state.

Graphs See Appendix.
Discussion of results and conclusion
From this experiment, we know the amount the spring stretches plotted against the weight added to the hanger gives a straight line that goes through the origin. This means that the extension of a spring is directly proportional to the stretching force applied to it. We can get the value of k for that spring by two ways. From experiment we get two value of the K that is KS and KT. I think these two values can not be approximately the same because there are friction of air and the geography in real life, but we omit them in the experiment. That causes the percentage difference between KS and KT

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