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Question

A vertical spring stretches $$3.9 \ cm$$ when a $$10-g$$ object is hung from it. The object is replaced with a block of mass $$25\ g$$ that oscillates up and down in simple harmonic motion. Calculate the period of motion.

Solution
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The spring constant is found from
$$k= \dfrac{F_s}{x}- \dfrac{mg}{x} = \dfrac{(0.010\ kg)(9.80 \ m/s^2)}{3.9 \times 10^{-2} \ m}= 2.5 \ N/m$$
When the object attached to the spring has mass $$m= 25 \ g$$, the period of oscillation is
$$T= 2\pi \sqrt{\dfrac{m}{k}} = 2 \pi \sqrt{\dfrac{0.025 \ kg}{2.5 \ N/m}}= \boxed{0.63 \ s}$$

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