Question

29. A particle performs a linear S.H.M. along a line
20√2 cm. find at what distance from the mean
position its K.E. is equal to its P.E. (Ans. x = =
10 cm)​

Answer 1

Answer:

Given:

Amplitude = 20√2 cm

To find:

Distance from mean position where potential energy and Kinetic energy will be same.

Formulas used:

$1. \: KE = \frac{1}{2} m { \omega}^{2}( {a}^{2} - {x}^{2} )$

$2. \: PE = \frac{1}{2} m { \omega}^{2} {x}^{2}$

Calculation:

As per the question;

PE = KE

$= > \frac{1}{2} m { \omega}^{2}( {a}^{2} - {x}^{2} ) = \frac{1}{2}m { \omega}^{2} {x}^{2}$

$= > {a}^{2} - {x}^{2} = {x}^{2}$

$= > 2{x}^{2} = {a}^{2}$

$= > x = \frac{a}{ \sqrt{2} }$

$= > x = \frac{20 \sqrt{2} }{ \sqrt{2} }$

$= > x = 20 \: cm$

So final answer is 20 cm from mean position.

Answer 2

$\huge{\colorbox{red}{Solution:}}$

$\huge{\pink{\underline{\underline{\mathfrak{Answer:}}}}}$

$\green{Mean \:position\: = 20 cm}$

$\huge{\pink{\underline{\underline{\mathfrak{Explanation:}}}}}$

$\green{As\:we\:know\:that,}$

$\orange{\begin{lgathered}1. \: KE = \frac{1}{2} m { \omega}^{2}( {a}^{2} - {x}^{2} ) \\\end{lgathered}}$

$\orange{\begin{lgathered}2. \: PE = \frac{1}{2} m { \omega}^{2} {x}^{2} \\\end{lgathered}}$

$\green{According\:to\: the\: question,}$

$\blue{Potential\:Energy = Kinetic\:Energy}$

$\blue{\begin{lgathered}= > \frac{1}{2} m { \omega}^{2}( {a}^{2} - {x}^{2} ) = \frac{1}{2}m { \omega}^{2} {x}^{2} \\\end{lgathered}}$

$\blue{\begin{lgathered}= > {a}^{2} - {x}^{2} = {x}^{2} \\\end{lgathered}}$

$\blue{{\begin{lgathered}= > x = \frac{a}{ \sqrt{2} } \\\end{lgathered}}}$

$\blue{\begin{lgathered}= > x = \frac{20 \sqrt{2} }{ \sqrt{2} } \\\end{lgathered}}$

$\green{=>x=20cm}$

$\red{Therefore,\: mean\: position\: is\: from \:20 cm.}$

$\pink{\underline{\underline{\mathfrak{Links\: of\:this\: Question→}}}}$

https://brainly.in/question/15199091