energy shm hard

Question

Fraction of time particle in SHM spends between x = -A/2 and x = A/2 is approximately:

RankGuru · Accepted Answer

Let the motion be $x = A \sin(\omega t)$. 
- The total period is $T = 2\pi / \omega$.
- To reach $x = A/2$: $A/2 = A \sin(\omega t) \implies \omega t = \pi/6$.
- To reach $x = -A/2$: $\omega t = -\pi/6$.
- Time spent in this range within one cycle: $\Delta t = 2 	imes \frac{\pi/6}{\omega} = \frac{\pi/3}{\omega}$ (one crossing) $	imes 2$ (one full period involves two crossings) = $\frac{2\pi}{3\omega}$.
- Fraction $= \frac{2\pi/3\omega}{2\pi/\omega} = \frac{1}{3}$. 
- *Correction*: The question asks for the region between $-A/2$ and $A/2$. In one full period, the particle crosses this region twice. Total time in region is $1/6 + 1/6 = 1/3$ of the cycle. Wait, the option says $1/3$ is correct. Option '2/3' was previously listed - let's re-verify. 
Actually, the phase angle for $-A/2$ to $A/2$ is $60^\circ$ ($-\pi/6$ to $\pi/6$). $60/360$ is $1/6$ for one pass, so $1/3$ for a full cycle. If $1/3$ is an option, it is the correct answer. (Note: The solution text previously mentioned $2/3$, likely a mistake).

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