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

Answer:

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

$\tt P.E=\dfrac{1}{2}\times k\times x^2$

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

Explanation:

Given:

• A particle executing SHM motion

To Find:

• The expression for K.E
• The expression for P.E
• The expression for total energy and to show that total energy is conserved
• Factors on which the total energy depends

Solution:

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,

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

Hence,

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

Also,

ω² = k/m

k = ω²m

Substituting this we get,

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

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,

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

W = k x²/2

$\boxed{\tt P.E=\dfrac{1}{2}\times k\times x^2 }$

Now total energy of the particle is given by,

T.E = K.E + P.E

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

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

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

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,

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

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

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

Answer 2

$\large{ \textsf{\textbf {\pink {➽ \: \: Answer \: :-}}}}$

Acceleration of the particle , performing S.H.M is given by α=−ω2y

where ω is the angular velocity, and y is the displacement of particle.

now, workdone by particle = F.dy

as we know, acceleration and displacement are in opposite directions in case of S.H.M

so, W=−mω2ydy

where m is the mass of the particle.

W=−mω2∫ydy

W=−21mω2y2

so, potential energy = -W

=21mω2y2

we know, ω=2πη

so, P.E=2π2η2my2 _____(1)

velocity of particle , v=ωAcosωt

or, v=ωA2−y2

so, kinetic energy of particle, K.E=21mv2

hence, K.E=21mω2(A2−y2)

but ω=2πη