What factors affect the rate at which my spring Oscillates?
Contents Contents Page 1 Brainstorm Page 2 Hypothesis Page 3 Plan: Diagram and List Of Apparatus Page 5 Method Page 6 Fair Test Page 7 Safety Page 8 Results: Table Of Results Page 9 Graph Page 10 Interpretation Page 13 Evaluation Page 14 Appendix Page 15 Brainstorm Hypothesis I think that the more weight you put on the spring, the more time it will take to make ten oscillations. To explain this I will start off at the beginning. According to Hooke's Law1 the extension is directly proportional to the force loaded onto it. The graph below shows the load and the extension. The line is straight meaning x=y or extension ? load. I think that if you doubled the weight you added to the spring then the extension of the spring will double as well2. I think that this will only happen, though, up until a certain point, because after that the spring will not revert to its original shape3. This point is called the elastic limit. The graph below shows the elastic limit of a spring. As you can see, at a certain point Hooke's law fails to work4. It starts off as E ? L until it reaches the elastic limit where the spring extends more than it would if E ? L. This is where the spring starts to stretch out of shape and will not go back to it's original state. As my investigation is into the
Investigate the way in which extension depends on the tension for rubber.
Brendan Lee Tension and Extension Aim: To investigate the way in which extension depends on the tension for rubber. Before I begin to do this experiment I need to know a little more about elastics. I got my self an elastic band and placed it over my forefinger on each hand. I gradually increased the tension of which I was applying. The original length of the elastic band was 3 cm, but when stretched to its furthest length, it had a length of 21 cm. This meant that it had an extension of 18 cm. The band could not stretch any further than this. If I had exerted even more tension the band would have snapped. I also noticed that after being stretched a few times, and then compared to an exact sized band that had not been stretched, that it did not return to its original shape. It had increased in size by a small amount. However if I only stretched the band a little bit each time, it would return to its original size. When tension is applied to the elastic band, the band automatically begins to repel that force that is stretching it, and when released the band moves back to its original position. When the band is stretched something is pulling the band in the opposite direction, a force. Now the band is stretched it has the potential to do work. We know this as if we released the tension on it the elastic band would pull its self back to its original size and shape. The band
Calculating the value of "g" (Gravitational field strength) using a mass on a spring
Calculating the value of "g" (Gravitational field strength) using a mass on a spring Gravity affects all things that have mass and therefore must affect how much a mass placed on a spring will extend. Measuring the time period and extension of a mass on a spring for vibrations should enable us to calculate a value for g. Using the following formula will help us to do this: Formula 1 T=2?Vm/k g (gravitational field strength) affects the spring constant - k in the formula F=ke and because F = weight = mg. Therefore mg = ke and m/k = e/g. We can now change formula 1 to the following: T=2?Ve/g If we rearrange the above formula so that the subject is T2 we should get the formula below: T2 = 4?2 e g Measuring T would allow us to calculate T2 (The time period - to calculate measure the time it takes for a certain number of oscillations and then divide it by the number of oscillations) and e would allow us to plot a graph and, according to the formula if we take the gradient of the line of best fit it will be equal to: 4?2 g We can then work out g, the gravitational field strength. g= 4?2 _ Gradient The graph that will be plotted will be T2 against e (time period2 against extension) and I expect that it will be similar to the following sketch: I predict that the gravitational field strength I calculate will be quite close to the 9.8N/Kg that is taken to be g
Refractive index by tracing light rays
Physics Laboratory Report Refractive index by tracing light rays Date of experiment: 24/3/2009 Aim of experiment: The objective of this experiment is: . to find out the refractive index of a glass block by Snell's Law 2. to find out the refractive index of water by Snell's Law Theory: In Part A of this experiment, the angle of incidence (i) and the angle of refraction (r) between air and a glass block were measured. As stated by Snell's Law, the following relation can be concluded: nair sin i = nglass sin r As nair = 1, sin i / sin r = nglass where nair and nglass are the refractive indexes of air and the glass block respectively. Hence, by the above equation and measurements of i and r, the refractive index of the glass block can be determined. In Part B of this experiment, the angle of incidence (i) between water and the glass block were measured. As stated by Snell's Law, the following relation can be concluded: nglass sin i = nwater sin r By substituting r =90o, nwater = nglass sin i where nwater is the refractive index of water. Hence, by the above equation, as the refractive index of water is found and i is measured, the refractive index of water can be determined. Apparatus: Rectangular glass block x 1 Pins x 4 Drawing board x 1 Protractor x 1 A4 paper x 2 Beaker of water x 1 Procedures: Part A:
Hooke's Law / Young's Modulus - trying to find out what factors effect the stretching of a spring.
Hooke's Law / Young's Modulus I am trying to find out what factors effect the stretching of a spring. Things, which might affect this, are: · Downward force applied to spring. · Spring material. · Length of spring. · No. of coils in spring. · Diameter of spring material. · Cross sectional area of spring. I have chosen to look at the effect of the weight applied, as it is a continuous variation. I predict that the greater the weight applied to the spring, the further the spring will stretch. This is because extension is proportional to load and so if load increases so does extension and so stretching distance. Extension = New length - Original length To see if my prediction is correct I will experiment, and obtain results using Hookes Law. He found that extension is proportional to the downward force acting on the spring. Hookes Law F=ke F = Force in Newtons k = Spring constant e = Extension in Meters My method of experimentation will be to use a clamp stand and boss clamp to suspend a spring from. A second boss clamp will hold in place a metre rule starting from the bottom of the spring to measure extension in mm. I will then add weights to the spring and measure extension. Before deciding on the range of experimentation I carried out a preliminary test to find the elastic limit of the springs we had. To do this I added weights to the spring until it
Investigating Hooke's Law
Investigating Hooke's Law Aim The aim of my coursework is to investigate and achieve a clear understanding into whether Hooke's law is true and to what extend in which it works and why. Prophecy Hooke's law states that if we add the same sized mass on to a spring its length should increase by a regular amount. For example when you double mass the extension should double. This should work until a spring reaches its elastic limit. The elastic limit of a spring is when the weight (stress) is too much and causes the spring to be permanently deformed and it does not return to its original length .The amount of deformation, as a fraction of the original size, is called strain. Elasticity is the property and the name given to a material that resumes its original size and shape after having been compressed or stretched by an external force. The elastic limit of a spring is determined by the molecular structure of the material of the actual spring. When a force is applied to the spring creating stress within the material, the molecular distances change and the material becomes deformed. Below the elastic limit, when the applied force is removed, the molecules return to their balanced position, and the elastic material goes back to its original shape. Beyond the elastic limit, the applied force separates the molecules to such an extent that they are unable to return to their
Investigating Hooke's Law into thin wires.
INVESTIGATING HOOKE'S LAW INTO THIN WIRES AIM To find out weather or not Hooke's Law applies to thin wires. THEORY This is an investigation into Hooke's Law and how it operates in relation to the thin wires we are going to be testing. the basic equation for Hooke's Law is F=KX. What this basically means is that the force is directly proportional to the extension. This is shown in Fig 1.1. In the darkly shaded area the material is elastic and thus the Law applies. However, if the external force is too strong, the material can become permanently deformed thus meaning that Hooke's Law on longer applies. This is represented in Fig 1.1 by the lightly shaded area. ================================================================ The elastic limit - also shown in Fig 1.1 - of a material is determined by the molecular structure of the material. The distance between molecules in a stress-free material depends on the balance between the molecular forces of attraction and repulsion. When an external force is applied, creating stress within the material, the molecular distances change and the material becomes deformed. If the molecules are tightly bound to each other, even for a large amount of stress there will be little strain. If, however, the molecules are loosely bonded to each other, a relatively small amount of stress will cause a large amount of strain. Below the elastic
The Electro magnetic spectrum.
The Electro magnetic spectrum. By Steve Wyers 11cu Radio Waves Radio waves are made by various types of transmitter, depending on the wavelength. They are also given off by stars, sparks and lightning, which is why you hear interference on your radio in a thunderstorm. Radio waves are the lowest frequencies in the electromagnetic spectrum, and are used mainly for communications. Radio waves are divided into:- Long Wave, around 1~2 km in wavelength. The radio station "Atlantic 252" broadcasts here. Medium Wave, around 100m in wavelength, used by BBC Radio 5 and other "AM" stations. VHF, which stands for "Very High Frequency" and has wavelengths of around 2m. This is where you find stereo "FM" radio stations, such as "Galaxy 101" and "GWR FM". Further up the VHF band are civilian aircraft and taxis. UHF stands for "Ultra High Frequency", and has wavelengths of less than a metre. It's used for Police radio communications, military aircraft radios and television transmissions. Large doses of radio waves are believed to cause cancer, leukaemia and other disorders. Some people claim that the very low frequency field from overhead power cables near their homes has affected their health. Microwaves Microwaves are basically extremely high frequency radio waves, and are made by various types of transmitter. In a mobile phone, they're made by a transmitter chip and an antenna,
For this investigation I have been asked to find out how different masses on a spring effect the extension when the springs are in parallel, series and on a single spring.
Planning Aim For this investigation I have been asked to find out how different masses on a spring effect the extension when the springs are in parallel, series and on a single spring. Key factors Independent variables: * Extension of spring Dependent variable * Mass on spring(s) Controlled variable * Springs The range of readings that I am going to take will be from 0kg to 0.50kg this is because it will give me a good set of data to work with. I will increase the mass by half a kilogram each time. To make sure I get good accurate fair results I will repeat the process at least 3 times. When I do repeat the process I will make sure that I leave all the equipment as it is and not replace bits or add or remove components. To make sure that I don't have to replace any components, I will before I start the test make sure that all my equipment is working correctly and properly calibrated to the range of readings that I will take in the test. Prediction I predict that the extension of a spring will be proportional to its load during its elastic region and that when the load of the spring is doubled so will the extension of the spring; this can then be used to find the spring constant. If two springs are placed in series, I believe that the extension of the springs will be double the extension of a single spring with the same load (therefore will have half the
Investigating the Vertical Oscillations of a Loaded Spring
Jackson Heddy 29/04/02 Investigating the Vertical Oscillations of a Loaded Spring Aims: The aim of this investigation is to find the elastic constant of the spring under study. The elastic constant of material is always useful to know and in some cases vital. For instance the elastic constant of a bungee rope is vital. It will tell the operator of the ride by how much the bungee rope will extend, using Hookes law (F=kExt.), and so how far the rider will fall. Using this they can calibrate the ride so as the give the rider the best possible experience. So a way of finding out the elastic constant of a material is important in today's jumping-off-bridges kind of world. Because of this the aim of this experiment is to find out through experimentation the elastic constant of a material and find out if it is the same as what is stated in the specifications of the material. The object under investigation is a metal spring. Method: To find the elastic constant of the spring I will attach a mass (a weight with its mass measured on the scales) to it, then stretch it to a set amount from its normal location (this being where the attachment hook rests when no other forces are applied). Doing this I will then let go and time the oscillations of the spring (one oscillation meaning leaving the starting position, passing the rest position, reaching the aphelion, then returning