Question

A simple pendulum with a period of 2.0s has its length doubled. its new period is?

Answer 1

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.

Step-by-step 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,

`T_1=2\pi\sqrt[]{(L)/(g)}\text{ }\rightarrow\text{ (i)}`

The new period of the pendulum is given by,

`T_2=2\pi\sqrt[]{(2L)/(g)}\text{ }\rightarrow\text{ (ii)}`

On dividing the equation (ii) by equation (i),

`\begin{gathered} (T_2)/(T_1)=\frac{2\pi\sqrt[]{(2L)/(g)}}{2\pi\sqrt[]{(L)/(g)}} \\ =\frac{\sqrt[]{2L}}{\sqrt[]{L}} \\ =\sqrt[]{2} \\ \Rightarrow T_2=T_1\sqrt[]{2} \end{gathered}`

On substituting the known values,

`\begin{gathered} T_2=2.0*\sqrt[]{2} \\ =2.83\text{ s} \end{gathered}`

Final answer:

The new period of the pendulum is 2.83 s