A graph of speed against time has also been produced, which is non-linear. This graph shows that the ball gains speed from zero out of the throwers hand. This speed increases sharply, but begins to fall as the force of gravity acts on it. As the ball reaches its peak in displacement, the speed decreases to 0m/s at its highest peak of 1.057m. However, the ball gains speed once more as the force of gravity accelerates the ball towards the ground. The graph shows this as the curve inclines steeply from 0m/s.
However, if tangents are drawn on this graph, the gradients will calculate the acceleration. This is because acc. = change of speed / change of time
Therefore, the gradient/acceleration from the initial throw = 4 / 0.05 = 80 m/s 2
The deceleration as the ball reaches the amplitude = -2.4 / 0.2 = -12 m/s 2
The final acceleration as the ball reaches the floor = 4 / 0.05 = 80 m/s 2
Part 2
Method
The procedure was repeated with the ball moving as a projectile. Rulers were used to measure the horizontal displacement and vertical displacement.
Results
Looking at a non linear graph G3, it shows the motion of the thrown squash ball in the air. It has a projectile motion where it reaches a peak in vertical displacement from the origin. Once reaching the peak, the squash ball is forced downwards by gravity and accelerating at the rate of gravity. Thus, as horizontal displacement increases, the vertical displacement increases until a peak of 0.34m in height. After, as horizontal displacement increases, the vertical displacement begins to decrease. The horizontal displacement increases throughout the throw.
The gradient at the start of the projection is, 0.37 / 0.16 = 0.432
The gradient at the end of the projection is, 0.15 / 0.4 = 0.375
These gradients are quite close which suggests that the motion of the ball at start and end are quite similar. The angles of elevation and de-elevation are also similar at 42 and 38 degrees.
On graph G4, it shows a non linear graph which increases sharply in speed once the ball is released from the throwers hand. As the ball reaches 0.04 seconds, the ball begins to lose it speed. Here, there is a decrease in speed until it reaches 0.2 seconds which is the half way point of the projection. The ball loses speed due to air resistance and the downward force of gravity. This is why the ball loses reaches its lowest speed (excluding the launch) at the peak of it vertical displacement at o.2 seconds. As gravity acts on the ball, it begins to increase in speed once more. This is because the ball accelerates with the force of gravity.
If tangents are drawn the gradients will represent the mean acceleration of the ball.
When the ball is released from the hand of the thrower, the acceleration is,
1.75 / 0.06 = 29.1 m/s 2
After, 0.04 seconds when the ball loses speed, the deceleration is,
1.05 / 0.1 = -10.5 m/s 2
As the ball approaches the ground after reaching the amplitude, the acceleration is,
1.975 / 0.1 = 19.75 m/s 2
Thus, in terms of acceleration, when the ball leaves the hand of the thrower, it accelerates highly in speed. However, along the journey towards the peak vertical displacement, it decelerates but once the amplitude is reached, gravity accelerates the ball towards the floor.
In a projectile, the vertical component increases at a rate of 9.8 m/s 2, which means that the rate of speed should increase as the ball falls further and further. This can be shown on the graph as the graph becomes steeper and steeper towards the end. The vertical component will also increase at a rate of – 9.8 m/s 2, so the rate of speed should decrease as the ball elevates higher and higher. This is shown on the graph as the graph becomes flatter and flatter towards the mid-point, where the ball reaches its peak in height.
Part 3
Method
A wooden beam is clamped to the top step of a ladder. Three springs were connected together and attached and attached to the beam. Then 400g of mass was attached to the bottom of the spring. The weights oscillated and the video was recorded.
Results
PTO
As shown on the graphs a sine wave is produced. Within the wave, it is noticeable that about 4 oscillations have been made by the spring. The time taken for 1 complete oscillation is called the period T. In this case the period T for 1 oscillation is about 1.28 seconds as shown on the graph. The number of oscillations per unit time is the frequency, f = 1 / T.
Therefore, in this case the frequency is, 1 / 1.28 = 0.78 Hz
Looking at the graph G5 the equilibrium position can be calculated as it will be the mid point of the maximum displacement. This equilibrium point is about 0.42m for the first oscillation. The oscillation as seen in the graph G5 can be described as simple harmonic motion. This is because the weights on the spring move about a fixed point such that its acceleration is proportional to its displacement from the fixed equilibrium point, and is directed towards the point.
Looking at the graph G6, it shows that as the spring falls and begins to oscillate, the speed begins to increase. As the spring reaches the equilibrium point, the velocity is at its maximum of about 2.2 m/s from the graph. When the spring fall downwards away from the equilibrium point, the speed decreases and eventually reaches 0 m/s at the very bottom of its stretch. Then, as the elastic potential lifts the weight back up towards the equilibrium point, the speed begins to increase again reaching its max. at the rest position. However, as the spring ascends to the highest point, the velocity decreases once more.
In terms of acceleration, the spring has a maximum acceleration when it is moving downwards towards the mid point. However, acceleration is at 0 when rest position is achieved. And deceleration begins as the weight reaches the bottom of its stretch. Acceleration increases again, as the spring ascends to rest point. Here acceleration decreases and is at 0m/s 2. The spring decelerates once more, as it ascends to its highest point. This spring shows the characteristics of simple harmonic motion.