Respuesta :
Answer:
9.82 m/s^2
Explanation:
T = 2 s, g = 9.8 m/s^2
T' = 1.99796 s
Let the acceleration due to gravity at new location is g'.
The formula for the time period of simple pendulum is given by
[tex]T = 2\pi \sqrt{\frac{L}{g}}[/tex] .... (1)
here, length of the pendulum remains same.
Now at the new location, let the time period be T'.
[tex]T' = 2\pi \sqrt{\frac{L}{g'}}[/tex] .... (2)
Divide equation (2) by equation (1), we get
[tex]\frac{T'}{T} = \sqrt{\frac{g}{g'}}[/tex]
[tex]\frac{1.99796}{2} = \sqrt{\frac{9.8}{g'}}[/tex]
[tex]0.99796 = {\frac{9.8}{g'}}[/tex]
g' = 9.82 m/s^2
tT9.82 m/s².
What is the time period of pendulum?
Pendulum is the body which is pivoted a point and perform back and forth motion around that point by swinging due to the influence of gravity.
The time period of a pendulum is the time taken by it to complete one cycle of swing left to right and right to left.
It can be given as,
[tex]T=2\pi \sqrt{\dfrac{L}{g}}[/tex]
Here, (g) is the gravitational force of Earth and (L) is the length of the pendulum.
The time period of the pendulum with a period of 2 s in one location g=9.80m/s2 can be given as,
[tex]2=2\pi \sqrt{\dfrac{L}{9.8}}\\L=0.996468\rm m[/tex]
Now, this pendulum is move to a new location where the period is now 1.99796 s. Thus, put the value in the above formula as,
[tex]1.99796=2\pi \sqrt{\dfrac{0.996468}{g}}\\g=9.82\rm m/s^2[/tex]
Thus, the acceleration due to gravity at its new location for the pendulum is 9.82 m/s².
Learn more about the time period of pendulum here;
https://brainly.com/question/3551146
