Question

The meter was originally defined so that the period of a meter-long simple pendulum would be exactly 2.00 second. (Was your measured value close to this?) Given the relationship, T^2 alpha L (T^2 is proportional to L) what would be the length of a simple pendulum, in centimeters, with a period of exactly one second?

Answer 1

Explanation:

Given that,

Initial length of simple pendulum, $L_1=1\ m$

Initial time period, $T_1=2\ s$

We need to find the length of the simple pendulum when the period is exactly 1 second.

$T_2=1\ s$

We know that the time period of simple pendulum is given by :

$T=2\pi \sqrt{\dfrac{L}{g}} \\\\T\propto \sqrt{L} \\\\\dfrac{T_1}{T_2}=\dfrac{L_1}{L_2}$

Put all values and find L₂

$\dfrac{T_1}{T_2}=\sqrt{\dfrac{L_1}{L_2}}\\\\L_2=\dfrac{T_2^2L_1}{T_1^2}\\\\L_2=\dfrac{1^2\times 100\ cm}{2^2\ s}\\\\L_2=25\ cm$

So, the length of the pendulum with a period of exactly one second is 25 cm.