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  • Draw a graph to show the variation of PE, KE and total energy of a simple harmonic oscillator with displacement.

Draw a graph to show the variation of PE, KE and total energy of a simple harmonic oscillator with displacement.
 

Answers (1)

Let us consider that a mass is lying on a horizontal frictionless surface, where spring constant = k.

If we displace the mass by distance A from its mean position, then it will execute SHM.

At this stretched position, P.E. of mass = \frac{1}{2} kA^{2}.

Now, at maximum stretch, i.e.,

At x = A, K.E. = 0

Now, at x = x\pm A

P.E. = total energy

               = \frac{1}{2} kA^{2}.

 Now let’s consider that the mass is back at its mean position, now the restoring force acting on the particle will be

P.E. = \frac{1}{2} kx^{2}

       = \frac{1}{2} m\omega ^{2}x^{2}

The restoring force constant of the oscillator isk = m\omega ^{2}

When x \pm A

P.E. = \frac{1}{2} m\omega ^{2}A^{2}

& K.E. = \frac{1}{2} mv^{2}

            = \frac{1}{2} m\omega ^{2} (A^{2} -x^{2})

Now, when x = \pm A,K.E. = 0

x

K.E.

P.E.

T.E.

0

\frac{1}{2} m\omega ^{2}A^{2}

0

\frac{1}{2} m\omega ^{2}A^{2}

+A

0

\frac{1}{2} m\omega ^{2}A^{2}

\frac{1}{2} m\omega ^{2}A^{2}

-A

0

\frac{1}{2} m\omega ^{2}A^{2}

\frac{1}{2} m\omega ^{2}A^{2}

Now, at displacement ‘x’, T.E. will be,

E = K.E. + P.E.

   = \frac{1}{2} m\omega ^{2}A^{2}

Hence, with displacement ‘x’, E is constant.

Posted by

infoexpert24

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