Question

Find the length of a second pendulum at a place where g = $10m{s^{ - 2}}$ take ( $\pi = 3.14$ ).

Answer 1

In order to solve this question we should know that the time required for the oscillations by the second’s pendulum is 2 seconds, then after we will put all the appropriate value that are given in questions in the formula of the time period of the pendulum and by doing this we will reach to the required solution.

Complete step by step answer:
For solving this question we need to utilize first that the oscillation time period of the second pendulum is 2 seconds:
Now as we already know that the relation between the length and the time period along with some more terms is given by the time period of the pendulum;
$T = 2\pi \sqrt {\dfrac{l}{g}}$
Here,
l is the length of the pendulum
g is acceleration due to gravity
T is the time period of the pendulum
Now since we know that taking square root in any expression is quite difficult than squaring it, ann there is a term here with square root so we will be squaring on both side:
Now the new expression will become in the following form:
${T^2} = 4{\pi ^2}\dfrac{l}{g}$
Now after putting all the values in given equation we will get:
$4 = 4{(3.14)^2}\dfrac{l}{{10}}$
Now substituting all the values on other side and calculating l we get;
$l = 1.01$ m
So, the length of the pendulum will be 1.01m.

Note: Pendulum, body suspended from a fixed point so that it can swing back and forth under the influence of gravity. Pendulums are used to regulate the movement of clocks because the interval of time for each complete oscillation, called the period, is constant. The formula for the period T of a pendulum is $T = 2\pi \sqrt {\dfrac{l}{g}}$ , where L is the length of the pendulum and g is the acceleration due to gravity.