Question

a. According to theory, the period T of a simple pendulum is T = 2????√ ???? ???? a. If ???? is measured as ???? = 1.40 ± 0.01 m, what is the predicted value of T?
b. Would you say that a measured value of T = 2.39 ± 0.01 ???? is consistent with the theoretical prediction of part (a)?

Answer 1

Answer:

a)         T = (2,375 ± 0.008) s , b) When comparing this interval with the experimental value we see that it is within the possible theoretical values.

Explanation:

a) The period of a simple pendulum is

         T = 2π √ L / g

Let's calculate

         T = 2π √1.40 / 9.8

         T = 2.3748 s

The uncertainty of the period is

         ΔT = dT / dL ΔL

         ΔT = 2π ½ √g/L   1/g  ΔL

         ΔT = π/g √g/L   ΔL

         ΔT = π/9.8 √9.8/1.4    0.01

         ΔT = 0.008 s

The result for the period is

        T = (2,375 ± 0.008) s

b) the experimental measure was T = 2.39 s ± 0.01 s

The theoretical value is comprised in a range of [2,367, 2,387] when we approximate this measure according to the significant figures the interval remains [2,37, 2,39].

When comparing this interval with the experimental value we see that it is within the possible theoretical values.

Answer 2

The time taken by the pendulum to complete an oscillation is called a period. The period of the given pendulum is 2.378 s.

The period of a simple pendulum:

$T = 2\pi \sqrt{\dfrac L g$

Where,

$T$ - period

$L$ - length of the pendulum = 1.4 m

$g$- gravitational acceleration = 9.8 m/s²

Put the values in the formula,

$T = 2\pi \sqrt{\dfrac {1.4}{9.8}}\\\\T = 2.3748 \rm \ s$

Therefore, the period of the given pendulum is 2.378 s.

Learn more about  the period of the simple pendulum:

https://brainly.com/question/14759840