Answer:
[tex]T\sqrt{2}[/tex]
Explanation:
First let us write out the formula connecting the simple parameters in a system of simple pendulum
[tex]\\ T=2\pi \sqrt{\frac{L}{g} }\\[/tex]
Where
T= period of oscillation,
L=length of the pendulum
g= acceleration due to gravity
from the above formula, it is obvious that Mass has no effect on the period(T).
Again let see the relationship that exist between the period(T) and the length(L).
let assume [tex]2\pi \frac{1}{\sqrt{g}} = k[/tex] , then we can have
[tex]T= k\sqrt{L}\\[/tex]
We can conclude that the period is directly proportional to the square-root of the length.
we can vary the constant K to have
[tex]\frac{T_{1} }{\sqrt{L_{1}}}=\frac{T_{2} }{\sqrt{L_{2}}} =...=\frac{T_{n} }{\sqrt{L_{n}}}[/tex]
Now back to the question, if the length was doubled, we have the
[tex]L_{2}=2L_{1}[/tex]
using the varying equation, we can substitute values and arrive at
[tex]\frac{T_{1} }{\sqrt{L_{1}}}=\frac{T_{2} }{\sqrt{2L_{1}}} \\[/tex]
carrying out simple operation, we arrive at
[tex]\frac{T_{1}*\sqrt{2}*\sqrt{L_{1} }}{\sqrt{L1} }=T_{2}\\[/tex]
Finally we have
[tex]T_{2}=T_{1} \sqrt{2} \\[/tex]
hence we can conclude that the When the length is doubled, the period is increase by a factor of root 2
