Question

The particle executing simple harmonic motion has a kinetic energy \[{{K}_{0}}{{\cos }^{2}}\omega t\]. The maximum values of the potential energy and the total energy are respectively.
\[\begin{align}
  & \text{A}\text{. }{{K}_{0}}\text{ and }{{K}_{0}} \\
 & \text{B}\text{. }0\text{ and }2{{K}_{0}}\text{ } \\
 & \text{C}\text{. }\dfrac{{{K}_{0}}}{2}\text{ and }{{K}_{0}} \\
 & \text{D}\text{. }{{K}_{0}}\text{ and }2{{K}_{0}} \\
\end{align}\]

Answer 1

Hint: We can solve the equations of the simple harmonic motion for the kinetic and the potential energy to find the total energy or we can simply find the answer by looking at its graph and by the given value of the kinetic energy and by applying simple physics laws.

Formula used:
\[\begin{align}
  & P.E.=\dfrac{1}{2}k{{x}^{2}} \\
 & K.E.=\dfrac{1}{2}m{{v}^{2}} \\
 & x(t)=A\cos (\omega t-\phi ) \\
 & v(t)=-A\omega \sin (\omega t-\phi ) \\
 & T.E.=P.E.+K.E. \\
\end{align}\]

Complete answer:
As it is given that kinetic energy is the cosine function therefore the maximum value of cosine can be 1 and hence the maximum value of kinetic energy can be given as
\[K.E{{.}_{\max }}={{K}_{0}}\text{ (As maximum value of co}{{\text{s}}^{2}}\omega t)\]
And the energy graph for the simple harmonic motion is given as
From the graph we can see that the total energy is constant and where potential energy is zero or the kinetic energy is maximum, in the actual case all the potential energy is converted into kinetic energy due to the law of conservation.

At \[P.E.=0\], \[K.E{{.}_{\max }}={{K}_{0}}\text{ }\]

Therefore maximum value of total energy is given as

\[T.E=K.E{{.}_{\max }}={{K}_{0}}\text{ }\]

When \[K.E.=0\], the total energy is given as

\[\Rightarrow T.E=P.E.\]

As the total energy of the simple harmonic motion is constant therefore we can say that the maximum value of potential energy will be
\[P.E.={{K}_{0}}\]
Hence, the maximum potential energy and maximum total energy will be \[{{K}_{0}}\text{ and }{{K}_{0}}\].

So, the correct answer is “Option A”.

Additional Information:
The potential energy (P.E.) and kinetic energy (K.E.) for the simple harmonic motion is given as
\[P.E.=\dfrac{1}{2}k{{x}^{2}}\text{ and }K.E.=\dfrac{1}{2}m{{v}^{2}}\]
And the value of displacement (x) and velocity (v) as function of time (t) is given as

 \[x(t)=A\cos (\omega t-\phi )\text{ and }v(t)=-A\omega \sin (\omega t-\phi )\]
And the total energy is given as

\[T.E.=K.E.+P.E.\]

Substituting value of K.E. and P.E.

\[T.E.=\dfrac{1}{2}m{{v}^{2}}+\dfrac{1}{2}k{{x}^{2}}\]

Now substituting value of x and v we get
\[\begin{align}
  & \implies

T.E.=\dfrac{1}{2}m{{(-A\omega \sin (\omega t-\phi ))}^{2}}+\dfrac{1}{2}k{{(A\cos (\omega t-\phi ))}^{2}} \\
 & \implies

T.E.=\dfrac{1}{2}m{{A}^{2}}{{\omega }^{2}}{{\sin }^{2}}(\omega t-\phi )+\dfrac{1}{2}k{{A}^{2}}{{\cos }^{2}}(\omega t-\phi ) \\
\end{align}\]
Where \[\omega =\sqrt{\dfrac{k}{m}}\]
\[\begin{align}
  & \implies

T.E.=\dfrac{1}{2}m{{A}^{2}}\left( \dfrac{k}{m} \right){{\sin }^{2}}(\omega t-\phi )+\dfrac{1}{2}k{{A}^{2}}{{\cos }^{2}}(\omega t-\phi ) \\
 & \implies

T.E.=\dfrac{1}{2}k{{A}^{2}}{{\sin }^{2}}(\omega t-\phi )+\dfrac{1}{2}k{{A}^{2}}{{\cos }^{2}}(\omega t-\phi ) \\
 & \implies

T.E.=\dfrac{1}{2}k{{A}^{2}}\left[ {{\sin }^{2}}(\omega t-\phi )+{{\cos }^{2}}(\omega t-\phi ) \right] \\
\end{align}\]
As \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\], therefore the above equation becomes
\[T.E.=\dfrac{1}{2}k{{A}^{2}}\]
Hence, we can see that the total energy is not dependent on x or v, it depends on only amplitude.

Note:
When applying formulas or laws directly we should be careful as the notation for the parameters can be different. Here \[{{K}_{0}}\] is the amplitude and it can be mistaken as the spring constant. Same applies with the graph, study of graphs or interpretation should be made carefully if we can’t interpret the graph, equation should be considered.