Question

Deduce the expressions for the kinetic energy and potential energy of a particle
executing S.H.M. Hence obtain the expression for total energy of a particle performing S.H.M and show that the total energy is conserved. State the factors on which total energy depends.​

Answer 1

$\begin{gathered} \\ \Large{\bf{\pink{\underline{Answer \::}}}} \\ \end{gathered}$

We know that kinetic energy is given by,

K.E = 1/2 × m × v²

where m is the mass and v is the velocity of the body.

But we know,

,

$\tt v=\pm \omega\sqrt{A^2-x^2}v$

Hence,

$\tt K.E=\dfrac{1}{2} \times m\times \omega^{2}\times (A^2-x^2)K.E=$

Also,

ω² = k/m

k = ω²m

Substituting this we get,

$\boxed{ \red{\tt K.E=\dfrac{1}{2} \times k\times (A^2-x^2)} }$

Now deducing the equation for potential energy. Here work done by the particle is stored as potential energy.

dW = -f dx

dW = -(-kx) dx

dW = kx dx

Integrating this we from position 0 to x get,

get,

$\displaystyle \sf \int\limits {dW}=\int\limits^x_0 k{x} \,$

W = k x²/2

$\boxed{ \red{\tt P.E=\dfrac{1}{2}\times k\times x^2 } }</p><p>$

Now total energy of the particle is given by,

T.E = K.E + P.E

$\tt { \red{T.E=\dfrac{1}{2}\:k\times (A^2-x^2)+ \dfrac{1}{2}\: k\times x^2 }}$

$\tt{ \red{T.E=\dfrac{1}{2}\:k(A^2-x^2+x^2)}}$

$\boxed{ \red{\tt T.E=\dfrac{1}{2}\:k\:A^2} }$

Here k and A are constants for a given SHM motion, therefore the total energy is conserved.

Factors on which total energy depends:

From the above expression,

$\tt \red{T.E=\dfrac{1}{2}\:k\:A^2 }$

$\tt \red{T.E=\dfrac{1}{2}\: m\: \omega^{2} \:A^2}$

Hence total energy depends upon the mass, angular velocity and amplitude of the particle.

Answer 2

$\sf\huge\bold{Answer:}$

Consider a particle of mass M performing linear SHM with amplitude A.

The restoring force acting on the particle is f = - kx

where, k = force constant

and x = the displacement of the particle from its mean position

① Kinetic energy :

At distance x from the mean position the velocity is

$\sf\bold\purple{:⟶v = \omega \: \sqrt{A {}^{2} - x {}^{2} } }$

where,$\sf\bold\purple {:⟶\omega = \sqrt{ \frac{k}{m} }}$.The Kinetic Energy (KE) of the particle is

$\sf\bold\purple{:⟶ KE = \frac{1}{2} mv {}^{2} = \frac{1}{2}m \omega {}^{2} \: ({A}^{2} - x {}^{2})}$

$\sf\bold\purple{ :⟶KE = \frac{1}{2} k \: ({A}^{2} - x {}^{2})...(i)}$

If the phase of particle at an in.stant is t is $\sf \theta \: = \omega \: t + x$ , where a is initial phase , its velocity at that inst.ant is

$\sf:⟶ v = \omega \:A cos( \omega \: t + x)$

and,its KE at that in.stant is

$\sf\bold\purple {:⟶KE = \frac{1}{2} mv {}^{2} = \frac{1}{2} \omega{}^{2} A {}^{2} \cos {}^{2} ( \omega \: t + x) }$

$\sf\bold\purple{:⟶ KE = \frac{1}{2} k A {}^{2} \cos {}^{2} ( \omega \: t + x) ...(ii) }$

Hence,KE varies with time as $\sf \cos 2\theta$

② Potential energy :

Consider a particle of mass m , performing a linear SHM along the path MN about the mean position O. Let the particle be at P, at a distance x from O.

The corresponding work done by the external agent will be dW = (-F)dx = kxdx. This work done is stored in the form of potential energy .The potential energy of the particle when its displacement from the mean position is x can be found by integrating the above expression from 0 to x.

$\sf\bold:⟶ PE = ∫ \: dW= \int\limits^x_0 {x} \: kx \\ = \frac{1}{2}kx {}^{2} ...(iii)$

The displacement of particle at an inst.ant t being

$\sf :⟶x = A \sin( \omega \: t + x )$

and PE at that in.stant is

$\sf\bold\purple{ :⟶PE = \frac{1}{2} kx {A}^{2} = \frac{1}{2} k { \: }^{2} \sin( \omega \: t + x) ...(iv)}$

Thus,PE varies with time as $\sf \sin {}^{2} \theta$

③ Total energy :

The total energy of the particle is equal to the sum of its potential energy and kinetic energy.

From equations (i) and (ii)

Total energy

E = PE + KE

$\sf\bold\blue{ = \frac{1}{2}kx {}^{2} + \frac{1}{2} k( {A}^{2} - x {}^{2} ) }$

$\sf \bold\blue{= \frac{1}{2}k {x}^{2} + \frac{1}{2} k {A}^{2} - \frac{1}{2} k {x}^{2} }$

$\bold\blue{ \therefore \: E = \frac{1}{2} k {A}^{2} = \frac{1}{2} m \omega {}^{2} {A}^{2} ...(v)}$

④ Total energy is conserved :

As m is constant , and $\sf \omega$ and a are constants of the motion, the total energy of the particle remains constant (or is conserved)

⑤ Factors on which total energy depends :

The total energy depends upon

• the mass (m)
• angular velocity $\sf \omega$
• amplitude of the particle ( A )