For first harmonic, our wavelength was 1.200 m, found by the formula λ=2L/n. In the second part we used tension and velocity to find mass density. For the first notch, with a tension of 10.3 N, the velocity was 86 m/s with a overall mass density of 0.018. Discussion In the lab we wanted to find the frequency at which resonance occurs on a stretched wire as well as look at the relationship between the frequency of vibration and tension and linear mass density of the wire. For the first procedure, we increased the frequency until we found resonance, and recorded frequencies and nodes to calculate wavelength.
Data: Procedure 1 m = .0657kg | M= .2426kg | y1= .059cm | Trial | p | y2 (m) | y2-y1 (m) | V (m/s) | Xvo (m/s) | 1 | 39⁰ | .145 | .087 | 1.31 | .315 | 2 | 38.5⁰ | .144 | .086 | 1.30 | .315 | 3 | 38.5⁰ | .143 | .084 | 1.28 | .311 | 4 | 38.5⁰ | .144 | .085 | 1.29 | .313 | 5 | 38.5⁰ | .144 | .085 | 1.29 | .313 | Our m is the mass of our ball and M is the mass of the pendulum just by itself. Our y1 is the distance from the table to the free hanging pendulum. To get V, we took the square root of 2g(y2-y1) which aided in calculating the initial velocity. The equation for Xvo was the mass of the ball plus the mass of the pendulum, divided by the mass of the ball all multiplied by the number we got for V. To get our y2 we measured the height of the pendulum
Discussion: In this experiment, a spring trigger launches a steel ball, acquiring a certain initial velocity. Then the ball collides with a wooden block, and together they reach a specific angel at height (Δh), called the angle indicator, and a final velocity. Our angle indicator reading was 0 degrees when the pendulum was hanging freely. Because of this, the maximum angel found from the collisions is easier to calculate, since we do not need to subtract the amount the pendulum is off from 0 degrees. After finding the initial estimate of the maximum angle, we loaded the launcher again and set the angle indicator to 3 degrees less than the angle reached in the previous step.
6) The procedure was then repeated for larger values of h, until T has passed its minimum value. (A smaller moveable knife edge must be used for values of h, as the axis of suspension lies on the narrow strip region of the pendulum) 7) A graph of T against h was then plotted and the value of h at which the period reaches a minimum should be estimated. 8) A graph of hT2 against h2 was then plotted and analysed to obtain a value for g. As a means of obtaining the value for g, the results were used to produce a straight line graph and the gradient of the graph used to find the value of g. The period of the pendulum is given by; T=2πω=2π(k2+h2)gh Squaring both sides of this equation gives; T2=4π2ghk2+h2 This re-arranges to give; hT2=4π2g×h2+4π2g×k2 Which is the form y=mx with y=hT2,
The difference is the buoyant force FB=Wout – Win. Finally we calculate percent error between the theoretical and experimental value for each shape: % error=(Measured – Theoretical)/Theoretical x 100%. In the next part of the lab, we compare the predicted and the measured capacity of a tuna can boat. The predicted capacity is calculated to be 0.183kg, and then we load the boat until it sinks and subtract 0.001kg to get the measured capacity. Finally, we analyze the errors in both parts of the lab by propagation by substitution and compare the theoretical-experimental values using errors.
Conclusion The purpose of this experiment was to see if your age mattered in how much lung capacity you would have. By measuring the distance around the balloon we would identify who has more lung capacity. My hypothesis was that I thought the older or younger you are the less lung capacity you have. Example- a one year old can’t blow a balloon as well as a ten year old yet a ninety year old can’t blow ass well as a ten year old to. My hypothesis was incorrect because we did the results of the data and did three different age groups which were: five through ten years old, twelve through twenty years old, and from thirty - nine through forty - three years old.
DCP: Circular Motion Aim: To investigate the effect of the radius of the circle on the time taken for 10 complete revolutions. Raw Data: Radius of Circle/cm (±0.1cm) | Time taken for 10 complete revolutions/s (±0.01s) | Average | | Trial 1 | Trial 2 | Trial 3 | | 25.0 | 4.74 | 5.18 | 4.76 | 4.89 | 30.0 | 5.34 | 5.02 | 4.89 | 5.08 | 35.0 | 5.89 | 5.63 | 5.74 | 5.75 | 40.0 | 6.43 | 6.18 | 6.11 | 6.24 | 45.0 | 6.63 | 6.55 | 6.50 | 6.56 | 50.0 | 7.04 | 6.88 | 6.95 | 6.96 | Figure 1 Calculations: -To find the average I added the results of the three trials together and divided by three. (e.g. : (4.74+5.18+4.76)/3=4.89) Processed Data: Radius of a circle/cm (±0.1cm) | Time taken for 10 complete revolutions/s (±0.01s) | Average time taken for 10 complete revolutions/s | Value | % Uncertainty | Trial 1 | Trial 2 | Trial 3 | Average Value | Absolute Uncertainty | % Uncertainty | 25.0 | ±0.4% | 4.74 | 5.18 | 4.76 | 4.89 | ±0.29 | ±5.9% | 30.0 | ±0.3% | 5.34 | 5.02 | 4.89 | 5.08 | ±0.26 | ±5.1% | 35.0 | ±0.3% | 5.89 | 5.63 | 5.74 | 5.75 | ±0.14 | ±2.4% | 40.0 | ±0.3% | 6.43 | 6.18 | 6.11 | 6.24 | ±0.19 | ±3.0% | 45.0 | ±0.2% | 6.63 | 6.55 | 6.50 | 6.56 | ±0.07 | ±1.1% | 50.0 | ±0.2% | 7.04 | 6.88 | 6.95 | 6.96 | ±0.08 | ±1.2% | Figure 2 Calculations: - For the absolute uncertainty I obtained the result by finding the difference between the mean and the result furthest from it: 5.18(furthest point from mean)-4.89(mean)= 0.29 Absolute uncertainty = ±0.2cm - For the percentage uncertainty I obtained the result by simply dividing the absolute uncertainty by the average and then multiplying by 100: (4.89/0.29)x100=5.9 Percentage uncertainty= ±5.9% Source of Uncertainties Source of uncertainties came mainly from the errors in equipment. The ruler used for the measuring of the radius had an limit of reading of 0.1cm, so
Displace the masses to cause oscillation 5. Measure the time taken for 10 oscillations, therefore get the time for 1 oscillation Set up the experiment as shown below: Results: length | Time | Ball size | 138.5 | 2.48 | S | 45.8 | 1.39 | S | 39.2 | 1.19 | S | 32.5 | 1.2 | M | 42.8 | 1.25 | M | 36.5 | 1.18 | M | 33.4 | 1.15 | M | 26.4 | 0.94 | L | 19.2 | 0.8 | L | 12 | 0.63 | L | Analysis: The graph of pendulum is shown on the page enclosed. The graph of the time taken for 1 oscillation and length of pendulum is a linear graph. The above table shows that time period of a pendulum varies with the length of the pendulum.The time taken for 1 oscillation is calculated by dividing the time for 10 oscillations. Question discussion: Why did we time 10 oscillations of the pendulum?
The total risk score is 4.14, the greatest relative or standardized difference between pretest and 3 month outcomes. This t ratio has a statistical significance of 0.05 - the least acceptable value for statistical significance. Also the larger the t ratio, the smaller the observed p value and increased odds of being able to reject the null hypothesis. 3. Which t-ratio listed in Table 3 represents the smallest relative difference between the pretest and 3 months?
What does this result mean? The t ratio of -0.65 represents the smallest relative difference between the pretest and 3 months outcomes. This ratio does not have an asterisk next to it in the table which according to the footnotes the asterisk is said to represent p < 0.05 the least stringent acceptable value for statistical significance. 4. What are the assumptions for conducting a t-test for dependent groups in a study?