Question

The potential energy of a simple harmonic oscillator when the particle is half way to its end point is:

• A. \(\frac 23\)E
• B. \(\frac 18\)E
• C. \(\frac 14\)E
• D. \(\frac 12\)E

Answer 1

The Correct Option is C

The potential energy (U) of a simple harmonic oscillator as a function of its displacement (x) from the equilibrium position is given by:
\(U =\frac 12kx^2\)
In the case of a simple harmonic oscillator, the maximum displacement (amplitude) is given by A, and when the particle is halfway to its endpoint, the displacement is \(\frac A2\).
So, the potential energy at this point is:
\(U = \frac 12k(\frac A2)^2\)

\(U = \frac 12k(\frac{ A^2}{4})\)

\(U = \frac 18kA^2\)
The total energy (E) of the simple harmonic oscillator is given by:
\(E = \frac 12kA^2\)
Now, to find the potential energy when the particle is halfway to its endpoint, we can substitute the expression for E:
\(\frac UE\)= \(\frac {\frac 18kA^2 }{ \frac 12kA^2}\)

\(\frac UE\)= \(\frac {\frac18}{\frac 12}\)

\(\frac UE\) = \(\frac 14\)

\(U\) = \(\frac 14E\)

So, the correct option is (C): \(\frac 14E\)