Newton's second law allows finding the result for the type of graph to find the gravity acceleration is:
g = [tex]\frac{slope}{\sqrt{L} }[/tex]
Oscillatory motion is a simple harmonic motion where the restoring force is proportional to the displacement.
Newton's second law gives a relationship between the net force, the mass and the acceleration of a body, in the attached we can see a free body diagram of the system, let's apply Newton's second law.
[tex]t_y - W =0 \\T_x = m a[/tex]
Since the movement is circular, the acceleration is centripetal.
[tex]a = \frac{v^2}{r}[/tex]
Let's substitute.
[tex]t_x = m \frac{v^2}{r}[/tex]
Let's use trigonometry to find the radius of the circle and the stress component.
sin θ = [tex]\frac{r}{L}[/tex]
r = L sin θ
sin θ = [tex]\frac{T_x}{T}[/tex]
cos θ = [tex]\frac{T_y}{T}[/tex]
Tₓ = T sin θ
[tex]T_y[/tex] = T cos θ
Let's substitute.
T sin θ = [tex]\frac{m v^2 }{L sin \theta }[/tex]
T sin² θ = [tex]\frac{m}{L} \ v^2[/tex]
From the other equation.
[tex]T_y= W \\T cos \theta = m g[/tex]
Let's write our system of equations.
T sin² θ = [tex]\frac{m}{L} \ v^2[/tex]
T cos θ = m g
We solve the system by squaring the second equation and dividing.
tan² θ = [tex]\frac{1}{g^2 L } \ v^2[/tex]
L g² tan² θ = v²
v = g [tex]\sqrt{L}[/tex] tan θ
Consequently, if the student makes a graph of the velocity versus the tangent of the angle, he obtains a straight line and the slope is:
slope = g [tex]\sqrt{L}[/tex]
From this expression you can calculate the acceleration of gravity.
[tex]g= \frac{slope}{\sqrt{L} }[/tex]
In conclusion using Newton's second law we can find the result for the type of graph to find the gravity acceleration is:
[tex]g = \frac{slope}{\sqrt{L} }[/tex]
Learn more here: brainly.com/question/17315536
