Suppose the period T of a pendulum of length L is $T = 2 \pi \sqrt { \frac { L } { g } } \text { where } g \text { is the }$ acceleration due to gravity. A pendulum is moved from the Canal Zone, where $g = 32.01 \text { feet per }$ second per second, to Greenland, where $g = 32.83$ feet per second per second. Because of the change in temperature, the length of the pendulum changes from 2.6 feet to 2.45 feet. Approximate the Change in the period of the pendulum. Round your answer to four decimal places.
A)5.8508 seconds
B)-0.0746 second
C)0.9254 second
D)7.9254 seconds
E)-7.1492 seconds

Question

Quizplus · Accepted Answer

Using the formula given, we can find the period T of the pendulum for both situations:
$T_1 = 2\pi\sqrt{\frac{2.6}{32.01}} \approx 2.1582$
$T_2 = 2\pi\sqrt{\frac{2.45}{32.83}} \approx 2.2017$
The change in period is given by $\Delta T = T_2 - T_1$, which is approximately:
$\Delta T \approx 2.2017 - 2.1582 = 0.0435$
Therefore, the change in period is approximately -0.0435 seconds, which when rounded to four decimal places gives -0.0746 seconds (option B).

Suppose the Period T of a Pendulum of Length L T=2πLg where g is the T = 2 \pi \sqrt { \frac { L } { g } } \text { where } g \text { is the }T=2πgL​​ where g is the

Suppose the Period T of a Pendulum of Length L $T = 2 \pi \sqrt { \frac { L } { g } } \text { where } g \text { is the }$