Question

If a simple pendulum has length \(l\) and time period \(T\) then a pendulum of length \(2l\) will have a time period of

• A. \(2\pi T\)
• B. \(\dfrac{1}{2\pi}T\)
• C. \(\sqrt{2}T\)
• D. \(\dfrac{1}{\sqrt{2}}T\)

Hint:

For a simple pendulum, the time period varies as the square root of the length: \[ T\propto \sqrt{l}. \] Doubling the length multiplies the time period by \(\sqrt{2}\).

Answer 1

The Correct Option is C

Step 1: Recall the formula for time period of a simple pendulum.
The time period of a simple pendulum is given by
\[ T=2\pi\sqrt{\frac{l}{g}} \] where \(l\) is the length of the pendulum and \(g\) is acceleration due to gravity.

Step 2: Write the formula for the new pendulum.
If the new length becomes
\[ l'=2l, \] then the new time period \(T'\) is
\[ T'=2\pi\sqrt{\frac{2l}{g}} \]

Step 3: Separate the square root terms.
\[ T' = 2\pi\sqrt{2}\sqrt{\frac{l}{g}} \]

Step 4: Use the expression for original time period.
Since
\[ T=2\pi\sqrt{\frac{l}{g}}, \] substitute into the above equation.
\[ T'=\sqrt{2}\,T \]

Step 5: Understand the dependence on length.
The time period of a pendulum is directly proportional to the square root of its length.
\[ T\propto \sqrt{l} \]

Step 6: Apply proportionality directly.
If length becomes twice, then
\[ T'=\sqrt{2}\times T \]

Step 7: Final conclusion.
Therefore, the time period of the pendulum of length \(2l\) is
\[ \boxed{\sqrt{2}T} \]
Hence, the correct answer is option (C).