Question

According to theory, the period T of a simple pendulum is T = 2pL=g (a) If L is measured as L = 1:40 0:01m; what is the predicted value of T ?
(b) Would you say that a measured value of T = 2:39 0:01s is consistent with the theoretical prediction of part (a)?

Answer 1

Answer:

a)the predicted value of T(ΔT) = (2.371 ± 0.008471)s

b) measured value of T = 2:39 ±0:01s is very close to part a So, we can say that result is consistent

Explanation:

Given data :

length of pendulum(L) =  1.40± 0.01 m

time of pendulum(T) =  $2\pi\sqrt{\frac{l}{g}}$

Where

g = 9.8 m/$s^{2}$

$\pi$ = 3.14

putting the value in time formula

T = 2×3.14$\sqrt{\frac{1.40}{9.8} }$= 2.371 s

uncertainty of time

T = $2\pi\sqrt{\frac{l}{g}}$

taking log on both sides

ln T = ln( $2\pi\sqrt{\frac{l}{g}}$)

In T = In 2$\pi$ + log $\sqrt{\frac{l}{g} }$ ( ln(axb) = ln a + ln b)

In T = In 2$\pi$ + log$(\frac{l}{g}) ^{\frac{1}{2} }$

In T = In 2$\pi$ + $\frac{1}{2}$ln $\frac{l}{g}$  (ln $a^{n}$ = n ln a)

In T = In 2$\pi$ + $\frac{1}{2}$ (ln l - ln g)

In T = In 2$\pi$ +$\frac{1}{2}$ ln l - $\frac{1}{2}$ln g

differentiating both side

$\frac{\Delta T}{T}$ = 0 + $\frac{1}{2}$ $\frac{\Delta l}{l}$ - 0

$\frac{\Delta T}{T}$ =  $\frac{1}{2}$ $\frac{\Delta l}{l}$

= $\frac{2.372}{2} \times \frac{0.01}{1.40}$

= 0.008471

Hence ΔT = (2.371 ± 0.008471)s

b) measured value of T = 2:39 ±0:01s is very close to part a So, we can say that result is consistent

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