Answer:
0.010 m
Explanation:
So the equation for a pendulum period is: [tex]y=2\pi\sqrt{\frac{L}{g}}[/tex] where L is the length of the pendulum. In this case I'll use the approximation of pi as 3.14, and g=9.8 m\s. So given that it oscillates once every 1.99 seconds. you have the equation:
[tex]1.99 s = 2(3.14)\sqrt{\frac{L}{9.8 m\backslash s^2}}\\[/tex]
Evaluate the multiplication in front
[tex]1.99 s = 6.28\sqrt{\frac{L}{9.8m\backslash s^2}[/tex]
Divide both sides by 6.28
[tex]0.317 s= \sqrt{\frac{L}{9.8 m\backslash s^2}}[/tex]
Square both sides
[tex]0.100 s^2= \frac{L}{9.8 m\backslash s^2}[/tex]
Multiply both sides by m/s^2 (the s^2 will cancel out)
[tex]0.984 m = L[/tex]
Now now let's find the length when it's two seconds
[tex]2.00 s = 6.28\sqrt{\frac{L}{9.8m\backslash s^2}}[/tex]
Divide both sides by 6.28
[tex]0.318 s = \sqrt{\frac{L}{9.8 m\backslash s^2}[/tex]
Square both sides
[tex]0.101 s^2 = \frac{L}{9.8 m\backslash s^2}[/tex]
Multiply both sides by 9.8 m/s^2 (s^2 will cancel out)
[tex]0.994 m = L[/tex]
So to find the difference you simply subtract
0.984 - 0.994 = 0.010 m
