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

Answer: (C)

$\frac 14$E

Answer 2

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): $\frac 14E$