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The potential energy of a particle of mass 14 Kg in motion along the x-axis is given by of U=8(1 - cos 12 x). the time period of the Particle for small oscillation $[\sin \theta=\theta]$ is $\frac{\pi}{Q} \sec$ The value of Q

Option: 1
4.54

Option: 2
2.31

Option: 3
5.47

Option: 4
6.29

Answers (1)

best_answer

$U=8(1-\cos 12 x) \\$

$F=-\frac{d u}{d x}=-8[12 \sin 12 x] \\$

$F=-8 \times 12 \sin 12 x \\$

$F=-96\, \sin 12 x \\$

$F=-96 \times 12 x \\$ ${[\because \sin \theta=\theta]} \\$

$F=1152 x \\$

$F=-1152 x \\$

$a=\frac{F}{m} \\$

$a=\frac{-1152 x}{14} \\$

$a=-82 \cdot 3\, x \\$

$\omega^2=82.3 \\$

$\omega=\sqrt{82.3}=9.072 \\$

$T=\frac{2 \pi}{\omega}=\frac{2 \pi}{9.072} \\$

$\text { or }=\frac{\pi}{\frac{9.072}{2}}=\frac{\pi}{4.54} \\$

$\text { or }=\frac{\pi}{4.54} \\$

$Q=4.54$

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Anam Khan

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