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The potential energy of a particle of mass $4 \mathrm{~kg}$ in motion along the $\mathrm{x}$ -axis is given by $\mathrm{U}=4(1-\cos 4 x) \mathrm{J}$ . The time period of the particle for small oscillation $(\sin \theta \simeq \theta)$ is $\left(\frac{\pi}{\mathrm{K}}\right) s.$ The value of $\mathrm{K}$ is___________.
Option: 1
2

Option: 2
-

Option: 3
-

Option: 4
-

Answers (1)

best_answer

$\mathrm{u =4(1-\cos 4 x) }$

$\mathrm{\frac{d u}{d x} =-4(-\sin 4 x) \times 4 }$

$\mathrm{=+16 \sin 4 x }$

$\mathrm{k_{\text {eff }} =\frac{d^2 u}{d x^2}=64 \cos 4 x }$
for small oscillation,

$\mathrm{x \rightarrow 0 }$

$\mathrm{4 x \rightarrow 0}$

$\mathrm{ \cos 4 x \rightarrow 1 }$

$\mathrm{K_{eff}=64}$

$\mathrm{\text { T } =2 \pi \sqrt{\frac{m}{k_{\text {cff }}}}=2 \pi \times \sqrt{\frac{4}{64}} }$

$\mathrm{T =\frac{\pi}{2} s}$

$\mathrm{\therefore K=2}$

The value of K is 2

Posted by

Irshad Anwar

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