Question

A pendulum is transported from sea-level, where the acceleration due to gravity g = 9.80 m/s2, to the bottom of Death Valley. At this location, the period of the pendulum is decreased by 5.00%

Answer 1

Answer:

$g'=10.78\ m/s^2$

Explanation:

The time period of a simple pendulum is given by :

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

L is length of the pendulum and g is acceleration due to gravity

At the bottom of Death valley, g = 9.8 m/s²

We need to find the value of g if the period of the pendulum is decreased by 5.00%.

T'=(T-0.05T)= 0.95 T

$T\propto \dfrac{1}{\sqrt{g} }\\\\\dfrac{T}{T'}=\sqrt{\dfrac{g'}{g}}\\\\\dfrac{T}{0.95T}=\sqrt{\dfrac{g'}{9.8}}\\\\\dfrac{1}{0.95}=\sqrt{\dfrac{g'}{9.8}}\\\\1.1=\dfrac{g'}{9.8}\\\\g'=10.78\ m/s^2$

So, the new value of acceleration due to gravity is $10.78\ m/s^2$.