Factors Affecting The Time Period of a Simple Pendulum
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Factors Affecting The Time Period of a Simple Pendulum
I plan to investigate the different characteristics of a simple pendulum and the factors affecting its time period. Any motion that repeats itself in equal intervals of time is called periodic motion. Simple harmonic motion (SHM) also is a periodic motion, but has some features: Â· It is periodic Â· Motion is to and fro along a straight line about an equilibrium (mean) position Â· Acceleration is directly proportional to displacement Â· Acceleration is directed towards the equilibrium position Therefore, simple harmonic motion is a periodic motion where acceleration is directly proportional to displacement, and is always directed towards the equilibrium position. Simple Pendulum What we call a simple pendulum is a string with a mass tied to the end of it. The mass is called a 'bob'. If we had kept the string with the bob attached to it at a fixed point and then we displace it from this equilibrium position, the movement of the bob we would see would be a to and fro movement where the bob would move in a straight arced line from left to right. The distance of the movement of the pendulum away from the equilibrium position is called its displacement. (Diagram 1 shows you the basic diagram of what was explained above) The three different types of oscillations are free, damped and fixed oscillations. Â· Free oscillations occur while the pendulum, is set to its displacement and is moving in its to and fro motion it doesn't experience any force that prevents it from continuing this motion. Such forces that prevent free oscillation are air resistance, etc. Â· Damped oscillations occur while the pendulum, is set to its displacement and is moving in its to and fro motion, experiences a force, or a medium that affects its motion. For example air resistance, water, etc. Â· A forced oscillation occurs while an object is used to force one or more pendulums into motion. An example of this is using a driving pendulum to control the displacement of a set of 4 pendulums, which move as a result of the driving pendulum being displaced. Another example is using a vibrating tuning fork to force a stretched string to vibrate and set the pendulum into motion. Time period The Time Period is the time that is required for one complete oscillation. (T) Frequency is the number of oscillations per second, which is measured in hertz (Hz). Angular Velocity the angle at which the radius vector subtends at the center in one second is called the angular velocity of the particle. Its unit is radians per second. That is, (rad s1) ====================================================================== Ï‰ = âˆ† (theta) / âˆ†t = 2Ï€ / t Key: Ï‰ = omega âˆ† (theta) = the angle which is covered âˆ†t = the time taken (Diagram 2 illustrates the simple pendulum using the formulae specified above) Suppose that you have a vector quantity, V. Theta is the angle dropped to the horizontal component. Any force can be resolved into a vertical component and a horizontal component. Cosine Theta = Adjacent / Hypotenuse Sine Theta = Opposite / Hypotenuse Cosine Theta = Adjacent / V Sine Theta = Opposite / V Adjacent = V (Cosine Theta) Opposite = V (Sine Theta) In the below diagram, the weight, 'Mg' (mass into gravity), can be resolved into two components. The first component 'Mg cosine' compensates for the tension in the string. 'Mg sine' is the component that provides the restoring force. The restoring force is required because it brings the mass back towards the equilibrium position. Initially it is a forced oscillation, but after two or three oscillations, the pendulum is in simple harmonic motion. We take a large number of oscillations, 10, because it reduces errors, as any errors that occur will be minimized. Factors Affecting Time Period Of A Simple Pendulum  * Length For the length of the pendulum, we consider the length from the suspension point of the string to the bottom of the bob, and then subtract the radius. While investigating the length, all other factors must be maintained constant for it to be a fair test. Therefore, the mass, amplitude and the acceleration due to gravity have to be constant. Since the amplitude and the mass will be constant, the bob will have to displaced by 5cm for the first test, and then so on accordingly for the other tests: Considering the law of energy that is, energy cannot be created or destroyed, we can say that all the potential energy at the ends will be transferred to the maximum kinetic energy that exists at the equilibrium position. Max P.E at extreme ends = Max K.E at the equilibrium position Mgh = Â½ mv2; cancel out mass and multiply by 2 on both sides 2gh = v2 âˆš2gh = v (Below is a diagram showing how the increase in length of string) Velocity is independent of the mass of the bob. If you use the same linear amplitude as before, when the length increases, the height of displacement reduces, the potential energy decreases, so kinetic energy decreases, velocity reduces and as the distance remains the same, Time Period increases. * Mass of the bob The mass of the bob could affect the time period, but the velocity is independent of mass, since we are maintaining the same linear amplitude, which means that it has the same speed and same distance, and so the time taken for an oscillation is the same. Therefore, a change in mass won't affect the time period of a simple pendulum. This is also proven by the fact that in the equation v = 2gh, the mass on both sides of the equation cancel each other out. * Amplitude The other factor that could affect the time period of the bob is its amplitude. To investigate this factor all other factors must remain constant. When the amplitude increases, the height of displacement increases, the potential energy increases, so kinetic energy increases, velocity increases and as the distance increases, but the Time Period remains the same. Â· Gravitational field strength the other factor that might affect the time period of the bob is gravitational field strength. To investigate this factor all other factors must remain constant. When the gravitational field strength increases, the potential energy increases, so kinetic energy increases, velocity increases and as the distance remains the same, but the Time Period decreases. Procedure: List of apparatus: * Different lengths of string * Digital stopwatch * Clamp stand * Meter ruler * Bobs (weights) of different masses The length of the pendulum is the length of the string from the suspension point of the string to the bottom of the bob, and then the radius of the bob has to be subtracted from this amount. Keeping the string at a longer length than that which is actually required, and then decreasing the length for each test can adjust length. For the investigations, I am planning to investigate a minimum of 5 length of string (40cm, 50cm, 60cm, 70cm, 80cm, 80cm, 90cm, etc.). I am planning to investigate a minimum of 3 different masses for the bob as well (400g, 800g, 1200g, etc.). And finally I am planning to investigate a minimum of 3 different amplitudes (6cm, 9cm, 12cm,). After the bob is set into motion, we must allow a few free oscillations because initially they are forced oscillations. If the starting point is taken as the extreme right an oscillation is said to be complete when the pendulum comes back to that specific side (right). (Below is a diagram illustrating this) The Bob moves from the position on the right, past the equilibrium then to the left, then back again past the equilibrium and then to the right side again. This is one complete oscillation. This is what the table for the different lengths would look like: Mass of bob = g Linear amplitude = cm Length (cm) Time for 10 oscillations Time period = t / 10 t1 (s) t2 (s) Avg. t = t1 + t2 / 2 40 50 60 70 80 90 100 110 This is what the table for the different masses would look like: Length = cm Linear amplitude = cm Mass (g) Time for 10 oscillations Time period = t / 10 t1 (s) t2 (s) Avg. t = t1 + t2 / 2 400 800 1200 1600 2000 2400 This is what the table for the different amplitudes would look like: Mass of bob = g Length = cm Amplitude (cm) Time for 10 oscillations Time period = t / 10 t1 (s) t2 (s) Avg. t = t1 + t2 / 2 6 9 12 15 18 Prior Test First I carried out a prior test using the same procedure as the procedure explained to see exactly how the experiment works, except for the fact that I was provided with different masses which were not the ones that I had predicted and I used different linear amplitudes and not the ones that I had predicted, and I had carried out three tests when investigating each factor, not 2 as I had planned. I used; Â· A string of length 50cm Â· A bob with mass 18g Â· A bob with radius 0.8cm The following is the table of results that I ended up with after carrying out the experiment: Amplitude (cm) Time for 10 oscillations Time period = t / 10 t1 (s) t2 (s) t3 (s) Avg. t = t1 + t2 / 3 5 14.16 14.20 14.14 14.17 1.417 10 14.26 14.12 14.13 14.17 1.417 15 14.17 14.16 14.22 14.18 1.418 Hence By looking at the above results, I can conclude that my predictions were correct because the amplitude doesn't affect the time period of the pendulum, when the mass and the length are kept constant. (Following is the graphical representation of these results on a graph.) ASHLEY. FERNANDES ================= 2. 3. 2004 SR 5 'B' PHYSICS CW FACTORS AFFECTING THE TIME PERIOD OF A SIMPLE PENDELUM OBTAINING EVIDENCE  After the prior test, I decided to carry out the experiments as described in the procedure, to analyse the length, mass and linear amplitude factors. BOB RADIUS (cm) MASS (g) BIG 1.3 65 MEDIUM 1 29 SMALL 0.8 18 These are the results that I had obtained when considering the length factor, where the mass was kept constant at 65g, the radius was 1.3cm and the linear amplitude was kept constant at 15cm: LENGTH (cm) TESTS (s) T = t/10 (s) First Second Thrid Average 30 11.22 11.22 11.28 11.24 1.124 40 12.9 13.04 12.92 12.95333 1.29533333 50 14.44 14.5 14.41 14.45 1.445 60 15.75 15.83 15.75 15.77667 1.57766667 70 17.22 17.03 17.25 17.16667 1.71666667 80 18.91 18.68 18.69 18.76 1.876 90 19.43 19.37 19.37 19.39 1.939 100 20.38 20.62 20.29 20.43 2.043 110 21.63 21.62 21.5 21.58333 2.15833333 120 22.03 22.19 22.31 22.17667 2.21766667 These are the results that I had obtained when considering the linear amplitude factor, where the mass was kept constant at 65g, the radius was 1.3cm and the length was also kept constant at 50cm: AMPLITUDE (cm) TESTS (s) T = t/10 (s) First Second Thrid Average 5 14.31 14.25 14.24 14.26667 1.42666667 10 14.34 14.36 14.32 14.34 1.434 15 14.44 14.5 14.41 14.45 1.445 These are the results that I had obtained when considering the mass factor, where the linear amplitude was kept constant at 15cm and the length was also kept constant at 50cm. The radius for each mass was 0.8cm (small bob), 1cm (medium bob) and 1.3cm (big bob) respectively: MASS (g) TESTS (s) T = t/10 (s) First Second Thrid Average Big 65 14.44 14.5 14.41 14.45 1.445 Medium 29 14.2 14.46 14.22 14.29333 1.42933333 Small 18 14.44 14.5 14.41 14.45 1.445 (Below are the graph done on plotted sheets to show length against time, amplitude against time and mass against time) ASHLEY. FERNANDES ================= 2. 3. 2004 SR 5 'B' PHYSICS CW FACTORS AFFECTING THE TIME PERIOD OF A SIMPLE PENDELUM ANALYSIS  Conclusions: 1. For the graph time period against length we suggest that it is not a linear graph, instead it is a curve. This is because as length increases, time period increases. For the same linear amplitude, as length increases the height through which the bob moves decreases. Hence when height decreases, kinetic energy will decrease, so also causing potential energy to decrease. Velocity therefore decreases. As velocity decreases, time period will increases. Therefore the prediction that I had made in planning is proven right. 2. The graph time period against mass does not affect time period. Mgh at extreme ends = Â½ mv at extreme ends V = âˆš2gh Velocity doesn't depend on mass. Distance remains the same, due to the linear amplitude being constant. So time period remains the same and therefore mass is not affected by time period. 3. Time period is not affected by change in amplitude. When linear amplitude increases, height through which the bob moves increases. When height increases potential energy and kinetic energy increases. As amplitude increases, the velocity and distance increases. Therefore time period is not affected. (Below are the specified graphs for the conclusions) We use the theory of Simple Harmonic Motion to find the relationship between the length and time squared. The relationship proves to be directly proportional. F = ma Ma = mg sine Î¸ A = g sine Î¸ Therefore, F = mg sine Î¸ The formula for angle in a sector, a = g Î¸, this is used to find Î¸ incase of it being small. a = g x A(1enght) / radius = g x BB1 / radius = g x  (x / length) In the above equation, 'BB1' represents the displacement and 'length' represents radius. Since 'a' (acceleration) opposes 'x', x will have a negative charge. Since 'g' and 'length' are constants, we can conclude that a is directly proportional to x. (feature of a simple harmonic motion) Therefore, a = Ï‰2x (where Ï‰ is Omega) So, Ï‰2 = a / x (where Ï‰2 is a positive constant) So, Ï‰2 = a / x Therefore, the time period, T, of a simple harmonic motion is: T = 2 Ï€ / Ï‰ T = 2Ï€ Ã· a / x T = 2Ï€ x x/a From the above equation, g x ( x / l) ; a = g x / l x / a = l / g Therefore, T = 2Ï€ l / g T2 = 4Ï€2 x l /g T2 = 4Ï€2 / g (l) Hence 4Ï€2 / g, is a constant Thus T2 is directly proportional to l L is directly proportional to T2 Therefore L directly proportional to T2 graph is needed. Following is a table of the length against time squared. This table is used to find the value for gravity. Length (m) T^2 (s) Gravitational val. 0.3 1.263376 9.37 0.4 1.677888444 9.41 0.5 2.088025 9.45 0.6 2.489032111 9.52 0.7 2.946944444 9.38 0.8 3.519376 9 0.9 3.759721 9.45 1 4.173849 9.46 1.1 4.658402778 9.32 1.2 4.918045444 9.63 9.399 Therefore the average gravity value obtained is, 9.399 â‰ˆ 9.40 I will hence plot a graph for length against time squared for the following values. Length (cm) T^2 (s) 30 1.263376 40 1.677888444 50 2.088025 60 2.489032111 70 2.946944444 80 3.519376 90 3.759721 100 4.173849 110 4.658402778 120 4.918045444 I am going to use the following formula to see if I obtain the right value for gravity from my graph. Gradient = T2 ' L g = 4Ï€2 x l / T2 Gravitational strength = 4Ï€2 x 1 / Gradient Gradient = (2.50 / 0.6)  (1.26 / 0.3) = 1.24 / 0.3 = 4.133 Gravitational strength = 4Ï€2 / 4.133 = 9.5520 'G' value from table 9.40 'G' value from graph 9.55 From the above statements, both values are around the value, 9.8 so I can conclude that my results were accurate. Quantitative Conclusions ======================== To confirm the accuracy of my results I will investigate further using quantitative tests. Sine Î¸ = A / L Î¸ = sin1 A / L Cosine = L1 / L = L  h / L (L Cosine) = L  h H = L  (L Cosine) V = Ï‰ x Î», where 'V' is the velocity, 'Ï‰' is Omega and 'Î»' is the amplitude Ï‰ = 2Ï€ / T Kinetic energy = Â½ mv2, where 'm' is the mass and 'v' is the velocity. Gravitational potential energy = M x g x H, where 'M' is the mass, 'g' is the gravity that is found from the table and the graph and 'H' is the height. Using the above formulas, I will find the height of displacement, velocity, kinetic energy, and potential energy. I will compare the values for kinetic energy and potential energy, and see if they are almost the same. If they are, then the experiment that I carried out was accurate. Following is the calculation as done for the 0.3m strings: = Sine1 A / L = Sine1 15 / 30 = 30Ëš H = L  (L Cosine) H = 30  (30 x cos30) H = 4.019cm = 0. 0401m V = Ï‰ x Î» V = 2Ï€ / T x 0.15m V = 0.838 m/s Kinetic energy = Â½ mv2 = Â½ x 0.065 x 0.8382 = 2.282293 x 102 J Gravitational potential energy = Mgh = 0.065 x 9.8x 0.0401 = 2.55437 x 102 J Length (L) (m) Height H = L  L Cosine (m) Velocity V = Ï‰ x Î» (m/s) Kinetic Energy K.E. = Â½ mv2 (J) Potential Energy P.E. = Mgh (J) 0.3 0.0401 0.838 0.02282293 0.0255437 0.6 0.9 From the above results I find that the values for kinetic energy and potential energy are quite similar, therefore I can conclude that my results were accurate. Looking at the table we see that as length increases, velocity decreases therefore the time period is increasing. ASHLEY. FERNANDES ================= 4. 3. 2004 SR 5 'B' PHYSICS CW FACTORS AFFECTING THE TIME PERIOD OF A SIMPLE PENDELUM EVALUATION  After analyzing the results that were obtained, I concluded that my investigation on the factors that effect the time period of a simple pendulum were correct. This can be established using the various values that were calculated from the results, including the value for the gravitational force acting on the bob. Moreover, once a pendulum is in simple harmonic motion, the potential energy it has at the extreme ends will be equal to the kinetic energy at its equilibrium position, due to none of its energy is lost; only instead changed. This was also indicated in the results obtained from the readings for the Potential energy that was calculated was similar to the value for the kinetic energy of the bob. From the tables in obtaining and analysis where we see the observed time period and calculated time period, we notice that there are slight differences in the values for the Time Period that were obtained and those that were calculated using the formula. (T = 2âˆ x / (L / g) [where g = 9.81 N/kg]) However, there is merely a difference of 0.01 of a second in most of the readings. Furthermore, both results show a general trend that as the length of the pendulum increases, the time period also increases. Taking this in consideration, I can conclude that my results, and the procedure itself, were accurate. I can conclude that though the method is reliable for investigation purposes, it is not very suitable for obtaining reliable results for calculation. This is because there is too much scope for human error, as in the recording of the time taken for ten oscillations (parallax error), in the measurement of the amplitude of the oscillations, and even in taking the length of the string for the pendulums created. To curtail this margin of error, perhaps pendulums of standard lengths could be used during the experiment, instead of measuring out different lengths. Also, for measuring the times for the various oscillations, Infra red sensors connected to computers could be used, so that the times taken for ten oscillations could be monitored and recorded precisely. Extended Investigation mg x I could investigate the oscillations of a spring mass system. This system is made up of a suspended spring that has a mass kept at its end. Pulling on this string increases the tension of the string hence a force that is countered by the weight of the mass, and the pulling force acting in the opposite direction. This is shown below: [IMAGE] [IMAGE] [IMAGE][IMAGE][IMAGE] [IMAGE] [IMAGE] [IMAGE] When the mass is pulled down, the tension in the string will increase. So, when the mass is released, it will move continuously up and down, oscillate, about its equilibrium position. The time period of this oscillation can then be found out using a calculation: In Figure 2, The force acting downwards = Mass of weight x Force of gravity on object Therefore, F = M x A Since the mass is at rest in Figure 2, the upward force (or tension) of the spring will be equal to the downward force (mg). This force can be given by the formula, F = k x e. 'K is the Spring Constant and 'E' is extension of the spring under the authority of mass. This force should be equal to the weight of the mass acting downwards. Therefore, Mg = K.E As shown in Figure 3, when the spring is pulled down, it is extended more by the distance (x). Considering this, the tension of the spring acting upwards once it is pulled further can be given by the formula, F = k (e + x) As the spring is being pulled downwards, there will be a resultant downward force (causing the spring to move downwards). This resultant downward force can be calculated using the formula: Resultant force 'F' (downwards) = Mg  k (e + x) Since Mg = K.E, the above equation can be formed as, F = K.E  k (e + x) Once lengthened, this equation will look as follows: F = K.E  K.E  KX F =  KX Therefore, F =  KX Since k is a constant, the relationship between the resultant downwards force and the extension can be written as, F Î±  x This states that as the resultant downward force increases, the extension of the spring (x) will also increase. Furthermore, since 'F = KX' and F = MA (Newton's 2nd Law), a new equation can be formed: MA = KX A = (k / m) x 'K' and 'M' are both constants, and so they can be replaced by 'Ï‰2,' which is a positive constant. Hence, Ï‰2 = k / m Therefore we get, Ï‰ = âˆš (k / m) For a SHM the time period = 2âˆ / Ï‰ Considering the previous equation, I can thus say, T = 2âˆ / âˆš (k / m) otherwise, T = 2âˆ x âˆš (m / k) In this fashion, the Time Period of the spring oscillations can be calculated. Also, the effect of the varying factors, such as using an object of greater mass, increasing the length of the spring, or even increasing the pulling force on the spring, on the Time Period of the oscillation, may perhaps be studied. How to Cite this Page
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