Question

Compare the kinetic energy and potential energy of a particle of a particle performing shm at the distance of x= A/2​

Answer 1

Answer:

18 Page no ans ahe maths

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Answer 2

Given,

Position of the particle x = $\frac{A}{2}$

where A = amplitude

To find,

The ratio of kinetic energy to potential energy.

Solution,

We know that,

The velocity of a particle performing S.H.M. at a distance of x from its mean position is;

v = ω $\sqrt{A^2-x^2}$

v = ω $\sqrt{A^2-(\frac{A}{2})^2 }$

$v^{2} = \frac{3}{4}$ Aω²

Now,

The kinetic energy(KE) of the particle performing S.H.M = $\frac{1}{2} mv^{2}$

                                                                                     = $\frac{1}{2} m \frac{3}{4} A^2$ω²

                                                                                     = $\frac{3}{8}$ mA²ω²

                                                                                     = $\frac{3}{8} k A^2$   (∴ω²=$\frac{k}{m}$)

The potential energy(PE) of a particle performing S.H.M is given by,

Potential Energy   = $\frac{1}{2} k x^2$      

                               = $\frac{1}{2} k \frac{A^2}{4}$

Now, Taking the ratio of KE to PE ;

⇒ $\frac{KE}{PE}$   = $\frac{\frac{3}{8}kA^2 }{\frac{1}{8}kA^2 }$

        = 3:1

Thus,

The ratio of KE to the PE of a particle executing SHM at a distance equal to A/2, the distance measured from its equilibrium position is 3:1.