### Related Formula In Simple Harmonic Motion: * In one full oscillation period T$T$, total distance = 4A$4A$, and net displacement = 0$0$. * In a quarter period fracT4$\frac{T}{4}$ starting from equilibrium, distance = A$A$, and displacement = A$A$. ### Core Logic Find the total number of cycles completed in the given time : textCycles = fractT = frac12.5text s2text s = 6.25text cycles = 6 text full cycles + 0.25text cycle$\text{Cycles} = \frac{t}{T} = \frac{12.5\text{ s}}{2\text{ s}} = 6.25\text{ cycles} = 6 \text{ full cycles} + 0.25\text{ cycle}$ ### Step 1: Evaluation of Distance and Displacement Calculate the total distance D$D$ covered : D = 6 times (4A) + 1A = 24A + A = 25A$D = 6 \times (4A) + 1A = 24A + A = 25A$ Given amplitude A = 1text cm$A = 1\text{ cm}$ , we get D = 25text cm$D = 25\text{ cm}$ . Calculate the net displacement d$d$: The first 6$6$ complete cycles return the particle to equilibrium with zero displacement. The remaining 0.25$0.25$ cycle (which is a quarter period fracT4$\frac{T}{4}$) moves it from the center to its maximum amplitude value : d = 1A = 1text cm$d = 1A = 1\text{ cm}$ Taking their ratio: fracDd = frac251 = 25$\frac{D}{d} = \frac{25}{1} = 25$ ### Pattern Recognition Fractional cycles modulate ratios abruptly. Always break the path index down into integer full cycles plus the final residual interval. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Oscillations