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 [tex]\frac{KE}{T} = \frac{3}{4}[/tex]
b
The fraction of the total energy of the system is potential energy [tex]\frac{PE}{T} = \frac{1}{4}[/tex]
Explanation:
From the question we are told that
The displacement of the system is [tex]e = \frac{a}{2}[/tex]
where a is the amplitude
Let denote the potential energy as PE which is mathematically represented as
[tex]PE = \frac{1}{2} * k* x^2[/tex]
=> [tex]PE = \frac{1}{2} * k* [\frac{a}{2} ]^2[/tex]
[tex]PE = k* [\frac{a^2}{8} ][/tex]
Let denote the total energy as T which is mathematically represented as
[tex]T = \frac{1}{2} * k * a^2[/tex]
Let denote the kinetic energy as KE which is mathematically represented as
[tex]KE = T -PE[/tex]
=> [tex]KE =k [ \frac{a^2}{2} - \frac{a^2}{8} ][/tex]
=> [tex]KE =k [ \frac{3}{8} a^2 ][/tex]
Now the fraction of the total energy that is kinetic energy is
[tex]\frac{KE}{T} = \frac{ \frac{3ka^2}{8} }{\frac{ka^2}{2} }[/tex]
[tex]\frac{KE}{T} = \frac{3}{4}[/tex]
Now the fraction of the total energy that is potential energy is
[tex]\frac{PE}{T} = \frac{\frac{k a^2}{8} }{\frac{k a^2}{2} }[/tex]
[tex]\frac{PE}{T} = \frac{1}{4}[/tex]
