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  • The potential energy of a particle of mass 9 kg in motion along the x-axis is given by <img alt="U=4(1+\sin 8 x)" src="https://entrancecorner.oncodecogs.com/gif.latex?U%3D4%281&plus;%5Csin

The potential energy of a particle of mass 9 kg in motion along the x-axis is given by U=4(1+\sin 8 x). The time period of the particle for small oscillation cos cos\, \theta = \theta is \frac{\pi }{k} sec. The value of K?

Option: 1

\frac{8}{3}


Option: 2

\frac{3}{8}


Option: 3

\frac{7}{3}


Option: 4

\frac{8}{4}


Answers (1)

best_answer

\begin{aligned} m & =9 \mathrm{~kg} \\ U & =4(1+\sin 8 x) \\ U & =4+4 \sin 8 x \\ F & =\frac{d U}{d x}=\frac{d}{d x}(4+4 \sin 8 x) \\ & =4(\cos 8 x) \cdot 8 \\ & =32 \cos 8 x \\ \cos \theta & =\theta \\ F_r & =32(8 x) \\ F r & =256 x \\ m w^2 x & =256 x \\ w^2 & =\frac{256}{m} \\ w & =\sqrt{\frac{256}{m}}\\ w & =\sqrt{\frac{256}{9}}\\ & \frac{2 \pi}{T}=\frac{16}{3} \\ & 6 \pi=16 T \\ & T=\frac{6 \pi}{16} \\ & T=\frac{3 \pi}{8} \\ & T=\frac{\pi}{8 / 3} \\ & \therefore K=\frac{8}{3} \end{aligned}

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Divya Prakash Singh

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