Question

For a particle doing SHM, Find expression for its Velocity and Potential energy.
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Answer 1

If an object is under oscillating motion and at any instant the force acting on it is proportional to its displacement from equilibrium position, then such an oscillating motion is called Simple harmonic motion (SHM).

Mathematically F(x) = -k×x....................(1)

where k is constant and negative sign indicates direction of force and direction of displacement are opposite.

 

Equation (1) can be rewritten by using the definition of force = mass times acceleration

begin mathsize 12px style negative k space x space equals space m fraction numerator d squared x over denominator d t squared end fraction space semicolon space h e n c e space fraction numerator d squared x over denominator d t squared end fraction plus k over m x space equals space 0 semicolon T h i s space d i f f e r e n t i a l space e q u a t i o n space h a s space s o l u t i o n space colon space x space equals space x subscript m cos left parenthesis omega t plus ϕ right parenthesis space............ left parenthesis 2 right parenthesis end style

 

where x is displacement, xm is amplitude of SHM, ω is angular frequency;  

ω= 2πf  where f is the frequency of oscillation and Φ is phase constant.

 

begin mathsize 12px style v e l o c i t y space o f space a n space o b j e c t space u n d e r space S H M space i s space fraction numerator d x over denominator d t end fraction equals space v subscript x space equals negative omega x subscript m sin open parentheses omega t plus ϕ close parentheses space................ left parenthesis 3 right parenthesis p o t e n t i a l space e n e r g y space U space equals 1 half k x squared space equals space 1 half space k space x subscript m superscript 2 space cos squared open parentheses omega t plus ϕ close parentheses space.............. left parenthesis 4 right parenthesis K i n e t i c space e n e r g y space K space equals space 1 half m v subscript x superscript 2 space equals space 1 half m omega squared x subscript m superscript 2 sin squared open parentheses omega t plus ϕ close parentheses............... left parenthesis 5 right parenthesis u sin g space left parenthesis 4 right parenthesis space a n d space left parenthesis 5 right parenthesis space w e space w r i t e space t o t a l space e n e r g y space K plus U space equals space 1 half k space x subscript m superscript 2 space space left parenthesis space w e space u s e d space t h e space r e l a t i o n space omega space equals square root of k over m end root space right parenthesis end style

hence total energy is constant

vishal

Answer 2

for a particul in SHM

V =

$w \sqrt{y {}^{2} - a {}^{2} }$

y=distance of particle from mean position

a=amplitude of partical

PE=

$\frac{1}{2} \times mw {}^{2}(y {}^{2} - a {}^{2} )$