Question

A mass of 0.250 ± 0.001kg is connected to the end of a light chord to make a pendulum. The length between the pendulum’s axis of rotation and the mass is measured to be 0.800±0.005m.

Auto Fit for: Latest | Angle Angle- A*sin(B*t+C)+D A: 0.07882 +/-0.0001835 B: 3.501 0.0004038 C: 5.017 - 0.004634 D: 0.08481 +/- 0.0001299 RMSE: 0.005803 rad 15 10 Time (s) Figure 1.0: The motion of the pendulum is recorded using a digital rotary motion sensor. The data are shown in red. The data are fit with a sinusoid function of the form: A*sin(B*t+C)+D thin black line. The resulting parameters obtained from the fit are displayed in the dialogue box.

3) Assume that the data presented in Figure 1.0 were collected using the same pendulum as the one described at the very top. How does the observed frequency in Exercise 2 compare to the theoretical frequency calculated in Exercise 1? Are the two frequencies compatible? Justify your answer.

Some helpful information:

*Theoretical frequency calculated in Exercise 1: 3.5 +/- 0.011 rad/s

*Observed amplitude and uncertainty in Exercise 2: 4.5 +/- 0.1

*Observed angular freuquency and uncertainty in Exercise 2: 3.501 +/- 0.0004038

Answer 1

we know that the solution of the exercise one which is described at the very top can be define with the dynamic equation

$\sum F_{neta}=-m*g*sin(\theta )=m*l*\frac{\mathrm{d}^{2} \theta }{\mathrm{d} ^{2}t}$

when $\theta <<<<1$ , $sin(\theta)\approx \theta$

then the equation change

$\frac{\mathrm{d}^{2} \theta }{\mathrm{d} ^{2}t}&plus;\sqrt{\frac{g}{l}}*\theta=0$

where $&plus;\sqrt{\frac{g}{l}}=\omega$    is the frequency now

it is possible that the rotational frequency meter solve the equation that defines the system through a numerical method and there is a numerical difference between the two results

we can see the solution of the differential equation is defined as:

$\theta =\theta _{0}*sin(\sqrt{\frac{g}{l}}*t)$

it is possible that the rotational frequency meter solve the equation that defines the system through a numerical method and there is a numerical difference between the two results
differential equation
this function is the solution of the equation is the same function that describes the movement of the number ascilatorio Figure 1 even if I have associated otroa as C that adds to the argument of the harmonic function and D displacing it upward, the latter It has to do with the initial conditions of study and the size of the amplitude at the initial moment of time

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2) although the initial conditions of exercise one is different than shown in Figure number one, the parameters describing the motion will have to be the same as they describe the same physical phenomenon that applies if the length between the pivot in the mass and weight they are the same for both exercise
then we have a coincidence frequency

ya que la frecuencia es un funcion de las magnitudes fisicas

$&plus;\sqrt{\frac{g}{l}}=\omega=3.501\, \frac{1}{s}$