Question 244: Oscillations

Question

The length of the simple pendulum is made 3 times the original length. If ' T ' is its original time period, then the new time period will be

Accepted Answer

We know, for a simple pendulum the time period is:
$ T = 2\pi \sqrt{\frac{L}{g}} $
where $L$ = length of pendulum and $g$ = acceleration due to gravity.
Step 1: Write the new time period
If the new length is $L' = 3L$, then the new time period is
$ T' = 2\pi \sqrt{\frac{L'}{g}} = 2\pi \sqrt{\frac{3L}{g}} $
Step 2: Compare with the original period
The original time period is
$ T = 2\pi \sqrt{\frac{L}{g}}. $
So,
$ T' = 2\pi \sqrt{\frac{3L}{g}} = 2\pi \sqrt{3}\sqrt{\frac{L}{g}} = \sqrt{3} \, T $
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Final Answer:
The new time period is
Option B: $\sqrt{3} \, T$

Answer

3 T

Answer

$\quad \sqrt{3} \mathrm{~T}$

Answer

$\frac{\mathrm{T}}{\sqrt{3}}$

Answer

$\frac{\mathrm{T}}{3}$

Oscillations

The length of the simple pendulum is made 3 times the original length. If ' T ' is its original time period, then the new time period will be

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