An object of mass 0.5, text{kg} is executing | vedclass

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(B) Given: Mass $m = 0.5\, 	ext{kg}$,Amplitude $A = 5\, 	ext{cm} = 0.05\, 	ext{m}$,Time period $T = 0.2\, 	ext{s}$.First,calculate the force const

Vedclass · Accepted Answer

$0.625$

Vedclass · Answer

(B) Given: Mass $m = 0.5\, 	ext{kg}$,Amplitude $A = 5\, 	ext{cm} = 0.05\, 	ext{m}$,Time period $T = 0.2\, 	ext{s}$.First,calculate the force constant $k$ using the formula $T = 2\pi \sqrt{\frac{m}{k}}$:$0.2 = 2\pi \sqrt{\frac{0.5}{k}}$$0.1 = \pi \sqrt{\frac{0.5}{k}}$$0.01 = \pi^2 \left(\frac{0.5}{k}ight)$$k = \frac{0.5 \pi^2}{0.01} = 50 \pi^2 \approx 50 	imes 9.87 = 493.5\, 	ext{N/m}$.Next,find the displacement $x$ at $t = \frac{T}{4}$ starting from the mean position $(\phi = 0)$:$x = A \sin(\omega t) = A \sin\left(\frac{2\pi}{T} \cdot \frac{T}{4}ight) = A \sin\left(\frac{\pi}{2}ight) = A = 0.05\, 	ext{m}$.Now,calculate the potential energy $PE$:$PE = \frac{1}{2} k x^2 = \frac{1}{2} 	imes (50 \pi^2) 	imes (0.05)^2$$PE = 25 \pi^2 	imes 0.0025 = 0.0625 \pi^2 \approx 0.0625 	imes 9.87 \approx 0.617\, 	ext{J}$.Rounding to the nearest provided option,we get $0.625\, 	ext{J}$.

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