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A body of mass $100 g$ attached to a spring executes SHM of period $2 s$ and amplitude $10 c m .$ If the body will take $\frac{x}{3} s$ to move from a point $5 c m$ below its equilibrium position to a point $5 c m$ above it, when it makes simple harmonic motion vertical oscillations. Then the value of x is ___ (take $g = 10 m / s^{2}$ )

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Detailed Solution

Period $2 \frac{π}{ω} = 2 s$ or $ω = π r a d / s,$ amplitude = $10 c m$

If the system is released, the equation of motion is $y = A \cos ω t$

$5 = 10 \cos π t_{1} w h e n y = 5 c m$

$- 5 = 10 \cos π t_{2} w h e n y = - 5 c m$

$π t_{1} = \cos^{- 1} (\frac{1}{2}) = \frac{π}{3} o r t_{1} = \frac{1}{3} s, π t_{2} = \cos^{- 1} (- \frac{1}{2}) = \frac{2 π}{3} o r t_{2} = \frac{2}{3} s$

Time interval = $t_{2} - t_{1} = \frac{2}{3} - \frac{1}{3} = \frac{1}{3} s$

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