Simple (really?) harmonic motion

Classical Mechanics Level 2

Calculate the length $(l)$ of a simple pendulum with a time period of $n \pi$ seconds.

Take $g$ is the acceleration due to gravity, which is constant in this case.

l = \dfrac {n^2g}{4}

l = \dfrac {n^2g}{8}

None of these

l =n^2g

l = \dfrac {n^2g}{2}

l = \dfrac {ng}{4}

1 solution

Akshat Sharda
Mar 25, 2016

Time period $(T)$ of a simple pendulum is,

$T=2\pi \sqrt{ \frac{l}{g} }$

Here $T=n\pi$ , $n\pi =2\pi \sqrt{ \frac{l}{g} } \Rightarrow \frac{n}{2}=\sqrt{ \frac{l}{g} } \Rightarrow \therefore l=\boxed{ \frac{n^2g}{4} }$

Good solution. Nice Ninja skills too.

Mehul Arora - 5 years, 2 months ago

Ninja skills!!!???

Akshat Sharda - 5 years, 2 months ago

Yeah. Don't let other post solutions, Ninja them :P

Mehul Arora - 5 years, 2 months ago

@Mehul Arora – What Ninja thing?

Akshat Sharda - 5 years, 2 months ago

(+1).... for the Ninja.... (+1)

Rishabh Jain - 5 years, 2 months ago

What are you guys talking about?

Akshat Sharda - 5 years, 2 months ago