Question
A particle is performing simple harmonic motion along x-axis with amplitude 4.0cm and time period 1.2s. What is the minimum time taken by the particle to move from x = +2cm to x = +4cm and back again?
✓
Answer
As $\text{x = a}\sin\omega\text{t}=\text{a}\sin\frac{2\pi\text{t}}{\text{T}}$ So, $\text{t}=\frac{\text{T}}{2\pi}\sin^{-1}\Big(\frac{\text{x}}{\text{a}}\Big),$ where a = 4cm At $\text{x}=2,\text{t}=\frac{\text{T}}{2\pi}\sin^{-1}\Big(\frac{2}{4}\Big)$$=\frac{\text{T}}{2\pi}\times\frac{\pi}{6}=\frac{\text{T}}{12}=\frac{1.2}{12}=\frac{1}{10}$
At $\text{x}=4,\text{t}=\frac{\text{T}}{2\pi}\sin^{-1}\Big(\frac{4}{4}\Big)$$=\frac{\text{T}}{2\pi}\times\frac{\pi}{2}=\frac{\text{T}}{4}=\frac{1.2}{4}=\frac{3}{10}$
Total Time taken $=\frac{3}{10}-\frac{1}{10}=\frac{2}{5}$
At $\text{x}=4,\text{t}=\frac{\text{T}}{2\pi}\sin^{-1}\Big(\frac{4}{4}\Big)$$=\frac{\text{T}}{2\pi}\times\frac{\pi}{2}=\frac{\text{T}}{4}=\frac{1.2}{4}=\frac{3}{10}$
Total Time taken $=\frac{3}{10}-\frac{1}{10}=\frac{2}{5}$
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