 # The Reflection Of The Definition Of A Simple Pendulum

Improved Essays
A pendulum follows a very simple type of oscillation called “isochronous”, which means it always takes the same time. Therefore these are oscillations that always repeat in the same time period maintaining the same this constant time even with a change in amplitude. A simple pendulum follows this pattern and it is why it has been such an important element in clocks. This can also be applied in different gravitational field strengths as the motion of the pendulum will always take the same time for one oscillation. Although there is a limit to the angle of a simple pendulum on the earth as if the amplitude is too much the pendulum will be slowed down by air friction.
When a pendulum bob is displaced from its equilibrium position, hanging vertically,
This can also be described using frequency per oscillation, which measures how many oscillations it does per second. Frequency is inverse of the period, f = 1/T. Also the period is inverse of the frequency, T = 1/f. We know that the horizontal component of the force is Tsinθ = -ma and the vertical component of the force is Tcosθ = mg. Therefore we can say that:
Tsinθ/Tcosθ= (-ma)/mg tanθ= (-a)/g
In our investigation we are going to keep the angle θ of the pendulum very small therefore we can say that the length L is going to be approximately the same for T. Also tanθ will approximately equal to sinθ. sinθ=(-a)/g x/l= (-a)/g ∴a=(-x)/l g

Using the laws of circular motion we can calculate the time period using the values for a we found above. w=2πf a=-(〖w)〗^2 x
Where w is the angular speed and f is the frequency. Using the equation above for a we can set it equal to the one we found previously to find T using f from angular speed.
(-x)/g=-(〖2πf)〗^2 x g/l=(〖2πf)〗^2 √(g/l)=2πf f=√(g/l)/2πf T=2π√(l/g)
We can also solve for g to find the acceleration due to gravity from our result by rearranging the equation.
g=(4π^2

## Related Documents

• Great Essays

When the spring is stretched, the spring force (Fs) tries to restore the spring to its original position. So the spring force Fs acts opposite to the displacement. We also know that force developed in a liner spring is directly proportional to the deformation (we only consider the case of liner spring). Since the displacement of the spring is opposite to the force Fs, work done by the spring force is negative. So we can write dU= -F_s.dx Integrating from position 1 to 2 ∫_1^2▒dU=-∫_(x_1)^(x_2)▒〖F_s dx〗 We know that for a liner spring Fs = kx.…

• 1828 Words
• 8 Pages
Great Essays
• Great Essays

To get a result as close to the ideal model as possible, the angle from the equilibrium point to the maxima should be significant small. So small that the amplitude, which has the shape of an arc, could be assumed to be a straight line. In that case, the oscillations could be treated as simple harmonic motion and the calculations would be able to give a realistic result. The period is obviously affected by the angle as well, since it takes longer time to complete one period if the angle is greater. As seen in table 1, our raw data has a fairly big spread even though the length of the string is constant, which is a result of different angles.…

• 1018 Words
• 5 Pages
Great Essays
• Improved Essays

Discussion From this experiment, it can be determined that the value of the coefficient of friction between the wire and the belt is 0.2782 (which is dimensionless). In addition, T1 becomes smaller (so the belt tension ratio becomes greater) when the angle of contact is increased, which can be explained by considering equation 2, as T2 cannot change for a given stationary mass. Comparing the experimental results to the theoretical predictions (made from the calculated value of μ, there is a relatively good agreement, especially when neglecting the πc result. The slightly higher gradient of the experimental results is likely due to an error (systematic) in the calibration of the load cell, causing it to become more inaccurate at higher tensions.…

• 729 Words
• 3 Pages
Improved Essays
• Great Essays

The equation of the line indicates that the period increased by 0.0003 seconds for every gram, this relationship between variables shows that there is hardly any change in period as the mass is increased in a pendulum. The trend was linear as the graph shows the equation for the line of best fit for the average being linear. This line is shown to cut through the period axis at 1.3593 seconds, as the gradient is so low, it is assumed that the period for all masses recorded was approximately 1.3593 seconds. The graph shows that as the mass of a pendulum ball increases, the period virtually isn’t affected. The table and graph both indicate low scatter between trials, this is seen on the graph as the different points for each increment of mass are close together.…

• 1881 Words
• 8 Pages
Great Essays
• Great Essays

The optimum stretching of TBC hydrogel was obtained at a draw ratio of 1.2 without any initiation of fracture in TBC. Stretching higher than 1.2 draw ratio showed initiation of fracture in the hydrogel for nearly all cases. On the other hand, lower cross-head speed was found to be the best for controlled stretching up to the maximum draw ratio. The cross-head speed was set to 0.03 mm/min based on the preliminary experiments. Lower cross-head speed gives sufficient time for maximum rearrangement and reduces the stress on nanofibrils.…

• 2188 Words
• 9 Pages
Great Essays
• Improved Essays

Which is reached by making r = 0. Therefore P_(L_MAX )= E^2/R_L with an ideal source. Thus for a completely resistive circuit the internal resistance of a DC source should be equal to zero for maximum power transfer to the load resistance. The derivation of the maximum power transfer theorem for resistive and non-linear impedance AC circuits. I – Phasor current (ωL or-1/ωC)_L and (ωL or-1/ωC)_r= X_L and X_r respectively I = (E_rms∠0°)/(Z+Z_r ) = (E_rms∠0°)/(r+R_L+j(X_r+X_L)) P_L= |I|^2 R_L =R_L/(〖(r+R_L)〗^2+〖(X+X_L)〗^2 ) 〖E_rms〗^2 2.1) The maximum power transfer when we have the choice to decide both the load resistance and reactance.…

• 771 Words
• 4 Pages
Improved Essays
• Improved Essays

The power consumption by the load is given as The VPV and Ri are constant, then Consider = 0, then . Hence load can achieve maximum power when Ri = Ro. The above explained simple method cannot be used to get the maximum power. To achieve the goal boost converter is used, which is shown in the Figure 4.9. Figure 4.9 Circuit diagram of boost converter employed in PV system The analysis is performed under the ideal condition.…

• 1370 Words
• 6 Pages
Improved Essays
• Improved Essays

Firstly, Newton’s first law states that an object at rest stays at rest and an object in motion stays in motion with the same speed and direction unless acted upon by an outside force. Nowadays, this law is also often referred to as the Law of Inertia. According to this law, the heavier object would have more inertia. Because it has more inertia, it also has a stronger resistance against changes in its state of motion. This larger amount of inertia offsets the larger amount of weight, so the two objects, although having different weights, would have the same rate of acceleration.…

• 632 Words
• 3 Pages
Improved Essays
• Superior Essays

The carpet can create large amounts of friction and the incline of the test course would cancel it out so that both the component of the gravitational force pushing the ball down the slope and the frictional force opposing it would be balanced leading to constant motion. As per Newton’s first Law of Motion, the Ping-Pong ball should maintain a constant velocity under these conditions. The rough corrugated sheet would produce too much friction and slow the object down as the frictional force would overcome gravity. The smooth corrugated sheet would create very little friction and would not oppose gravity, leading to acceleration. The other two cases are expected to…

• 1268 Words
• 6 Pages
Superior Essays
• Great Essays

The ruler that we used in this experiment had a range of 0-30 cm with a sensitivity of 0.05 cm. The ruler needs no calibration. Our other measuring device was the spring scale. The spring scale was used to measure the force of weight on each cylinder in newtons. The spring scale used in this experiment had a range of 0-2.5 newtons and a sensitivity of 0.025 newtons.…

• 935 Words
• 4 Pages
Great Essays