Respuesta :
Answer:
1) [tex] L \propto T^2 [/tex]
Using the condition given:
[tex] 2.205 m = K (3)^2 [/tex]
[tex] K = 0.245 \approx \frac{g}{4\pi^2}[/tex]
So then if we want to create an equation we need to do this:
[tex] L = K T^2[/tex]
With K a constant. For this case the period of a pendulumn is given by this general expression:
[tex] T = 2\pi \sqrt{\frac{L}{g}}[/tex]
Where L is the length in m and g the gravity [tex] g = 9.8 \frac{m}{s^2}[/tex].
2) [tex] T = 2\pi \sqrt{\frac{L}{g}}[/tex]
If we square both sides of the equation we got:
[tex] T^2 = 4 \pi^2 \frac{L}{g}[/tex]
And solving for L we got:
[tex] L = \frac{g T^2}{4 \pi^2}[/tex]
Replacing we got:
[tex] L =\frac{9.8 \frac{m}{s^2} (5s)^2}{4 \pi^2} = 6.206m[/tex]
3) [tex] T = 2\pi \sqrt{\frac{0.98m}{9.8\frac{m}{s^2}}}= 1.987 s[/tex]
Step-by-step explanation:
Part 1
For this case we know the following info: The length, l cm, of a simple pendulum is directly proportional to the square of its period (time taken to complete one oscillation), T seconds.
[tex] L \propto T^2 [/tex]
Using the condition given:
[tex] 2.205 m = K (3)^2 [/tex]
[tex] K = 0.245 \approx \frac{g}{4\pi^2}[/tex]
So then if we want to create an equation we need to do this:
[tex] L = K T^2[/tex]
With K a constant. For this case the period of a pendulumn is given by this general expression:
[tex] T = 2\pi \sqrt{\frac{L}{g}}[/tex]
Where L is the length in m and g the gravity [tex] g = 9.8 \frac{m}{s^2}[/tex].
Part 2
For this case using the function in part a we got:
[tex] T = 2\pi \sqrt{\frac{L}{g}}[/tex]
If we square both sides of the equation we got:
[tex] T^2 = 4 \pi^2 \frac{L}{g}[/tex]
And solving for L we got:
[tex] L = \frac{g T^2}{4 \pi^2}[/tex]
Replacing we got:
[tex] L =\frac{9.8 \frac{m}{s^2} (5s)^2}{4 \pi^2} = 6.206m[/tex]
Part 3
For this case using the function in part a we got:
[tex] T = 2\pi \sqrt{\frac{L}{g}}[/tex]
Replacing we got:
[tex] T = 2\pi \sqrt{\frac{0.98m}{9.8\frac{m}{s^2}}}= 1.987 s[/tex]
