Question

An object suspended from a spring vibrates with simple harmonic motion. Part A At an instant when the displacement of the object is equal to one-fourth the amplitude, what fraction of the total energy of the system is kinetic

Answer 1

Complete Question

An object suspended from a spring vibrates with simple harmonic motion.

a. At an instant when the displacement of the object is equal to one-half the amplitude, what fraction of the total energy of the system is kinetic?

b. At an instant when the displacement of the object is equal to one-half the amplitude, what fraction of the total energy of the system is potential?

Answer:

a

The fraction of the total energy of the system is kinetic energy  $\frac{KE}{T} = \frac{3}{4}$

b

The fraction of the total energy of the system is potential energy  $\frac{PE}{T} = \frac{1}{4}$

Explanation:

From the question we are told that

    The displacement of the system is  $e = \frac{a}{2}$

where a is the amplitude

     

Let denote the potential energy as PE  which is mathematically represented as

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

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

          $PE = k* [\frac{a^2}{8} ]$

 Let denote the total energy as T which is mathematically represented as

           $T = \frac{1}{2} * k * a^2$

Let denote the kinetic energy as  KE  which is mathematically represented as

      $KE = T -PE$

  =>     $KE =k [ \frac{a^2}{2} - \frac{a^2}{8} ]$

=>      $KE =k [ \frac{3}{8} a^2 ]$

Now the fraction of the total energy that is kinetic energy is  

       $\frac{KE}{T} = \frac{ \frac{3ka^2}{8} }{\frac{ka^2}{2} }$

       $\frac{KE}{T} = \frac{3}{4}$

Now the fraction of the total energy that is potential energy is  

      $\frac{PE}{T} = \frac{\frac{k a^2}{8} }{\frac{k a^2}{2} }$

      $\frac{PE}{T} = \frac{1}{4}$