If the length of a second's pendulum is doubled, what will be the time period (take the initial time period as 1 second)?

CONCEPT:

• Simple pendulum: When a point mass is suspended with the help of a string or rod of negligible mass and does the to and fro motion about its mean position is called a simple pendulum.
• For a simple pendulum, the time period of swing of a pendulum depends on the length of the string and acceleration due to gravity.

\(T = 2{{\pi }}\sqrt {\frac{{\rm{l}}}{{\rm{g}}}}\)

The above formula is only valid for small angular displacements.

Where, T = Time period of oscillation, l = length of the pendulum, and g = gravitational acceleration

CALCULATION:

Let initial length = l

\(Time\;period\;\left( T \right) = 2{\rm{\pi }}\sqrt {\frac{{\rm{l}}}{{\rm{g}}}}\)

The time period of a second's pendulum is 2 seconds, that is,

\(T= 2\pi\sqrt[]{\frac{l}{g}}=1\)

If the length is doubled, then the new time period (T*) will be

\(T^*= 2\pi\sqrt[]{\frac{2l}{g}}\\ \;\;\;\;\;=\sqrt{2}\times2\pi\sqrt[]{\frac{l}{g}}\\ \;\;\;\;\;=\sqrt{2}\times1\\ \;\;\;\;\;= 1.414 \;sec \)

So option 1.414 sec  is correct.