Question

When the displacement of a mass on a spring in simple harmonic motion is A/2 from the equilibrium position, what fraction of the total mechanical energy is kinetic energy?

Answer 1

Answer:

The ratio is  KE : TM  =  0.75

Explanation:

from the question we are told that

  The displacement of a mass on a spring in simple harmonic motion is A/2 from the equilibrium position

Generally the total mechanical  energy of the mass is mathematically represented as

        $TM = \frac{1}{2} * k * A^2$

Here  k is the spring constant  ,  A is the total displacement of the  the mass  from maximum  compression to maximum extension of the spring

Generally this total mechanical energy is mathematically represented as

        $TM = KE + PE$

=>     $KE = TM - PE$

Here the potential  energy of the mass is mathematically represented as

     $PE = \frac{1}{ 2} * k * [ x ]^2$

Here x is the displacement of the mass from maximum compression or extension of the spring to equilibrium position and the value is  

      $x = \frac{A}{2}$

So

     $PE = \frac{1}{ 2} * k * [ \frac{A}{2} ]^2$

So

      $KE = \frac{1}{2} * k * A^2 - \frac{1}{2} * k * [\frac{A}{2} ]^2$

=>    $KE = \frac{1}{2} * k * A^2 - \frac{1}{8} * k * A ^2$

=>    $KE = 0.375 * k * A^2$

So the ratio of  $KE : TM$ is  mathematically represented as

       $\frac{KE}{TM} = \frac{0.375 k A^2 }{0.5 k A^2}$

=>    $\frac{KE}{TM} = 0.75$