Oscillations PYQ Using Energy Concept

❓ Question

The position of a particle executing SHM is

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

Between $t=0$ and $t=T$ (where $T$ is the time period), the potential energy is minimum at

$$t=\frac{T}{2\beta}.$$

Find the minimum positive value of $\beta$.

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✍️ Short Explanation

For SHM,

$$U=\frac{1}{2}kx^2$$

Potential energy is minimum when displacement is zero:

$$x=0.$$

Given

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

therefore

$$\sin(\omega t)=0.$$

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🔷 Step 1 — Find the Instants of Minimum P.E.

$$\omega t=n\pi$$
$$t=\frac{n\pi}{\omega}.$$

Since

$$\omega=\frac{2\pi}{T},$$
$$t=\frac{n\pi}{2\pi/T} =\frac{nT}{2}.$$

Thus the minima of potential energy occur at

$$t=0,\ \frac{T}{2},\ T.$$

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🔷 Step 2 — Use the Given Condition

Given

$$t=\frac{T}{2\beta}.$$

For the minimum positive value of $\beta$, choose the largest valid time in $0<t\le T$ at which P.E. is minimum.

The largest such time is

$$t=T\mathrm{.}$$

Hence

$$\frac{T}{2\beta}=T.$$
$$\frac{1}{2\beta}=1$$
$$\boxed{\beta=\frac12}.$$

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✅ Final Answer

$$\boxed{\frac{1}{2}}$$

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🔷 JEE Trap Alert

A common mistake is taking the first non-zero minimum at

$$t=\frac{T}{2},$$

which gives $\beta=1$.

But the question asks for the minimum positive value of $\beta$. Since

$$\beta=\frac{T}{2t},$$

$\beta$ is minimum when $t$ is maximum, i.e., $t=T$.

Therefore,

$$\boxed{\beta_{\min}=\frac12}.$$

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