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Question Number 130978 by I want to learn more last updated on 31/Jan/21

Commented by I want to learn more last updated on 31/Jan/21

$The question start this way :$ $A spring of negligible mass is clamped vertically as shown in the$ $figure . A 2 kg . . . .$
Commented by mr W last updated on 31/Jan/21

$a n s w e r g i v e n i s w r o n g .$
Commented by I want to learn more last updated on 31/Jan/21

$Please help me sir . Thanks .$
Commented by I want to learn more last updated on 31/Jan/21

$Sir, when you are chanced, please help me .$
Commented by Tawa11 last updated on 14/Sep/21

$nice$
	Answered by mr W last updated on 31/Jan/21

	Commented by mr W last updated on 31/Jan/21

	$e n e r g y b a l a n c e p o i n t 0 a n d p o i n t 1 :$ $m g (h + Δ l) = \frac{1}{2} k {(Δ l)}^{2}$ $m = 2 k g$ $h = 0.25 m$ $Δ l = 0.3 m$ $2 \times 10 \times (0.25 + 0.3) = \frac{1}{2} k \times 0 . 3^{2}$ $\Rightarrow k = \frac{2200}{9} \approx 244.4 N / m$ $p e r i o d :$ $T = 2 π \sqrt{\frac{m}{k}} = 2 π \sqrt{\frac{2 \times 9}{2200}} \approx 0.568 s$ $y_{0} = \frac{m g}{k} = \frac{2 \times 10 \times 9}{2200} = \frac{9}{110} \approx 0.081 m$ $a m p l i t u d e :$ $A = Δ l - y_{0} = 0.3 - \frac{9}{110} = \frac{12}{55} \approx 0.218 m$ $y = - y_{0} + A \cos (\sqrt{\frac{k}{m}} t + π)$ $y = - \frac{9}{110} + \frac{12}{55} \cos (\sqrt{\frac{1100}{9}} t + π)$ $a t t = 0.5 s :$ $y = - \frac{9}{110} + \frac{12}{55} \cos (\sqrt{\frac{1100}{9}} \times 0.5 + π) \approx - 0.241 m$ $(a) p e r i o d T = 0.568 s$ $(b) d i s p l a c e m e n t a t t = 0.5 s : 0.241 m$
	Commented by I want to learn more last updated on 31/Jan/21

	$Wow, i really appreciate sir .$
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