Calculating Beta ($\beta$) for Maximum Potential Energy in SHM

The displacement of a particle in Simple Harmonic Motion (SHM) is given by:

$x(t) = A\sin(\omega t)$

Where $A$ is the amplitude and $\omega$ is the angular frequency.

The potential energy (PE) of an SHM oscillator is proportional to the square of its displacement:

$PE(t) = \frac{1}{2} k x(t)^2$

Substituting the expression for displacement:

$PE(t) = \frac{1}{2} k (A\sin(\omega t))^2 = \frac{1}{2} k A^2 \sin^2(\omega t)$

The potential energy is maximum when $\sin^2(\omega t)$ is maximum.

• The maximum value of $\sin^2(\omega t)$ is 1.
• This occurs when $\omega t = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \dots$, or generally $\omega t = (n + \frac{1}{2})\pi$, where $n$ is an integer ($n=0, 1, 2, \dots$).

Let's consider the first time this maximum occurs after $t=0$, which corresponds to $n=0$.

$\omega t = \frac{\pi}{2}$

$t = \frac{\pi}{2\omega}$

We know the relationship between the time period $T$ and angular frequency $\omega$ in SHM:

$\omega = \frac{2\pi}{T}$

Substitute this into the expression for $t$:

$t = \frac{\pi}{2 \left( \frac{2\pi}{T} \right)} = \frac{\pi T}{4\pi} = \frac{T}{4}$

The problem states that the maximum potential energy occurs at $t = \frac{T}{2\beta}$.

Equating the two expressions for the time $t$ when PE is maximum:

$\frac{T}{4} = \frac{T}{2\beta}$

By comparing both sides, we find:

$\frac{1}{4} = \frac{1}{2\beta}$

Solving for $\beta$:

$2\beta = 4$

$\beta = 2$