X-અક્ષ પર ગતિ કરતા 3 kg દળના પદાર્થનું સ્થાન | vedclass

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(A) આપેલ છે,$m = 3 \ kg$ દળના કણનું સ્થાન $x = 0.3 \cos (\omega t)$ છે.કણનો વેગ,$v = \frac{dx}{dt} = \frac{d}{dt} (0.3 \cos \omega t) = -0.3 \omega \s

Vedclass · Accepted Answer

$1 / 3$

Vedclass · Answer

(A) આપેલ છે,$m = 3 \ kg$ દળના કણનું સ્થાન $x = 0.3 \cos (\omega t)$ છે.કણનો વેગ,$v = \frac{dx}{dt} = \frac{d}{dt} (0.3 \cos \omega t) = -0.3 \omega \sin (\omega t)$.ગતિઊર્જા $K(t) = \frac{1}{2} m v^2 = \frac{1}{2} m (-0.3 \omega \sin \omega t)^2 = \frac{1}{2} m (0.09 \omega^2 \sin^2 \omega t)$.$t_1 = \frac{\pi}{6 \omega}$ સમયે,$K\left(\frac{\pi}{6 \omega}\right) = \frac{1}{2} m (0.09 \omega^2 \sin^2 \frac{\pi}{6}) = \frac{1}{2} m (0.09 \omega^2) \left(\frac{1}{2}\right)^2 = \frac{1}{2} m (0.09 \omega^2) \left(\frac{1}{4}\right)$.$t_2 = \frac{\pi}{3 \omega}$ સમયે,$K\left(\frac{\pi}{3 \omega}\right) = \frac{1}{2} m (0.09 \omega^2 \sin^2 \frac{\pi}{3}) = \frac{1}{2} m (0.09 \omega^2) \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{1}{2} m (0.09 \omega^2) \left(\frac{3}{4}\right)$.ગુણોત્તર લેતા: $\frac{K\left(\frac{\pi}{6 \omega}\right)}{K\left(\frac{\pi}{3 \omega}\right)} = \frac{1/4}{3/4} = \frac{1}{3}$.

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