Question

The length of second pendulum is decreased by 0.3 cm when it is shifted to chennai from london. If the acceleration due to gravity at London is 981 cm/sec^2, the acceleration due to gravity at chennai is :

Answer 1

Answer on Question #65622, Physics / Astronomy | Astrophysics

The length of second pendulum is decreased by 0.3 cm when it is shifted to chennai from london. If the acceleration due to gravity at London is 981 cm/sec^2, the acceleration due to gravity at chennai is :

Find: $g_2 - ?$

Given:

$\Delta l = 0.003\ \mathrm{m}$

$g = 9.81\ \mathrm{m/s^2}$

$T = 2\ \mathrm{s}$

Solution:

The period of simple pendulum:

$$T = 2 \pi \sqrt{\frac{l}{8}} (1)$$ $$\text{Of (1)} \Rightarrow \sqrt{\frac{l_1}{g_1}} = \frac{T}{2\pi} (2)$$ $$\text{Of (2)} \Rightarrow \frac{l_1}{g_1} = \frac{T^2}{4\pi^2} (3)$$ $$\text{Of (3)} \Rightarrow l_1 = \frac{T^2 g_1}{4\pi^2} (4)$$ $$\text{Of (4)} \Rightarrow l_1 = 0.981\ \mathrm{m} (5)$$

The length of simple pendulum in Chennai:

$$l_2 = l_1 - \Delta l \quad (6)$$

(5) in (6): $l_2 = 0.978\ \mathrm{m}$ (7)

$$\text{Of (1)} \Rightarrow \sqrt{\frac{l_2}{g_2}} = \frac{T}{2\pi} (8)$$ $$\text{Of (8)} \Rightarrow \frac{l_2}{g_2} = \frac{T^2}{4\pi^2} (9)$$ $$\text{Of (9)} \Rightarrow g_2 = \frac{4\pi^2 l_2}{T^2} (10)$$ $$\text{Of (10)} \Rightarrow g_2 = 9.78\ \mathrm{m/s^2}$$

Answer:

978 cm/s²

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