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A body of mass 100 g attached to a spring executes SHM of period 2 s and amplitude 10 cm. How long time is required for it to move from a point 5 cm below its equilibrium position to a point 5 cm above it, when it makes simple harmonic vertical oscillations (take g = 10 m/ $s^{2}$ )?

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a

0.6 s

b

1/3 s

c

1.5 s

d

2.2 s

answer is B.

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Detailed Solution

Period $= 2 \frac{π}{ω}$ = 2s (or) w = p rad/s, amplitude = 10 cm. If the system is released, the equation of motion is $y = A c o s ω t$ $\begin{array}{l} 5 = 10 c o s π t_{1} w h e n y = 5 c m \\ - 5 = 10 c o s π t_{2} w h e n y = - 5 c m \end{array}$

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

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