Given:
The period of a simple pendulum, T₁=2.0 s
The length of the pendulum is doubled.
To find:
The new value of the period of the pendulum.
Explanation:
Let us assume that, initially, the length of the simple pendulum was L. Then, its new length will be, 2L.
The period of a simple pendulum is given by the equation,
[tex]T_1=2\pi\sqrt[]{\frac{L}{g}}\text{ }\rightarrow\text{ (i)}[/tex]The new period of the pendulum is given by,
[tex]T_2=2\pi\sqrt[]{\frac{2L}{g}}\text{ }\rightarrow\text{ (ii)}[/tex]On dividing the equation (ii) by equation (i),
[tex]\begin{gathered} \frac{T_2}{T_1}=\frac{2\pi\sqrt[]{\frac{2L}{g}}}{2\pi\sqrt[]{\frac{L}{g}}} \\ =\frac{\sqrt[]{2L}}{\sqrt[]{L}} \\ =\sqrt[]{2} \\ \Rightarrow T_2=T_1\sqrt[]{2} \end{gathered}[/tex]On substituting the known values,
[tex]\begin{gathered} T_2=2.0\times\sqrt[]{2} \\ =2.83\text{ s} \end{gathered}[/tex]Final answer:
The new period of the pendulum is 2.83 s
