Question

Derive this formula.
$\tt T = 2\pi \sqrt{ \dfrac{l}{g} }$
Thank you. :)​

Answer 1

Explanation:

Explanation:

Let's consider that two consecutive positive integers are x & (x + 1).

$\begin{gathered}{\underline{\sf{\bigstar\: According \ to \ the \ given \ Question :}}}\\ \\\end{gathered}$

$\begin{gathered}:\implies\sf x^2 + \Big(x + 1 \Big)^2 = 365 \\\\\\:\implies\sf x^2 + x^2 + 1 + 2x = 365 \\\\\\:\implies\sf 2x^2 + 2x^2 = 365 - 1 \\\\\\:\implies\sf 2x^2 + 2x- 364 = 0 \qquad \bigg\lgroup\sf Taking \ 2 \ common \bigg\rgroup\\\\\\:\implies\sf x^2 + x - 182 = 0 \\\\\\\qquad\qquad\underline{\sf{\purple{\: Using \ splitting \ the \ Middle \ term \ method \ :}}}\\\\\\:\implies\sf x^2 + 14 x - 13x - 182 = 0\\\\\\:\implies\sf x( x + 14) - 13(x + 14) = 0\\\\\\:\implies\sf\pink{(x - 13) (x + 14) = 0}\\\\\\:\implies\sf x - 13 = 0 \\\\\\:\implies\boxed{\rm{\blue{\: x = 13}}} \\\\\\:\implies\sf x + 14 = 0 \\\\\\:\implies\boxed{\rm{\blue{\: x = -14}}}\end{gathered}$

$\bigstar★ Finding numbers$

First number (x) = 13

Second number (13 + 1) = 14[/tex]

⠀⠀⠀

$\begin{gathered}\therefore\underline{\textsf{Two positive consecutive numbers are \textbf{13 \& 14}}}. \\ \end{gathered}$

Answer 2

Derivation :

$\tt T = 2\pi \sqrt{ \dfrac{l}{g} }$

For a restoring force of any system in SHM

$\sf \bf F = - mg \: \sin \theta$

hence,

$\implies \sf \: ma = - \frac{m \: g \: x}{l}$

$\implies \sf \: \cancel{m}a = - \frac{ \cancel{m} \: g \: x}{l}$

$\implies \sf \: \bf {{a = - g \: \frac{x}{l} }\underline{ \: \: \: \: \: \: \: }(1)}$

Again,

we know

$\checkmark \: \bf \: \: { { a = - \omega ^{2} x}} \underline{ \: \: \: \: \: \: \: }(2)$

Comparing

$\sf - \frac{g}{l} x = { \omega}^{2} x \\ \\ \\ \implies \sf \omega = \sqrt{ \frac{g}{l} } \\ \\ \\ \implies \sf \:2 \pi n = \sqrt{ \frac{g}{l} } \\ \\ \\ \implies \sf \: \frac{1}{n} = T = 2 \pi \sqrt{ \frac{l}{g} } \\ \\ \\ \implies{ \underline{ \boxed{ \tt{T = 2 \pi \sqrt{ \frac{l}{g} } }}}}$

${\bf \: Note :}$

• $\: \tt \: This \: \: \omega \: \: is \: \: not \: \: angular \: \: velocity \: \: it's \: \: angular \: \: frequency. \: \: In \: \: physics\: \: both \: \: are \: \: having\: \: same \: \: symbol (\omega)$
• $\tt \: we \: \: use \: \:T = 2 \pi \sqrt{ \frac{l}{g} } \bigg(1 + \frac{1}{ {2}^{2} } \sin^{2} \frac{ \theta}{2} + \frac{1}{ {2}^{2} } . \frac{ {3}^{2} }{ {4}^{2} } { \sin}^{4} \frac{ \theta}{2} + \dots \bigg) \bf \: \: [when \theta > 4]$

$\small \tt \colorbox{aqua}{@StayHigh}$