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vladimir2022 [97]
3 years ago
5

A particle of mass m moves under an attractive central force F(r) = -Kr4 with angular momentum L. For what energy will the motio

n be circular, and what is the radius of the circle? Find the frequency of radial oscillations in the particle is given a small radial impulse.

Physics
1 answer:
docker41 [41]3 years ago
3 0

Answer:

Angular velocity is same as frequency of oscillation in this case.

ω = \sqrt{\frac{7K}{m} } x [\frac{L^{2}}{mK}]^{3/14}

Explanation:

- write the equation F(r) = -Kr^{4} with angular momentum <em>L</em>

- Get the necessary centripetal acceleration with radius r₀ and make r₀ the subject.

- Write the energy of the orbit in relative to r = 0, and solve for "E".

- Find the second derivative of effective potential to calculate the frequency of small radial oscillations. This is the effective spring constant.

- Solve for effective potential

- ω = \sqrt{\frac{7K}{m} } x [\frac{L^{2}}{mK}]^{3/14}

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The end point of a spring oscillates with a period of 2.0 s when a block with mass m is attached to it. When this mass is increa
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<h3>What is simple harmonic motion?</h3>
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  • Where, m is the mass of the body and k is the spring constant.
<h3>How to solve the problem?</h3>
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               T_1=2s\\m_1=m\\m_2=m+2kg\\T_2=3s

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              T_1=2\pi \sqrt{\frac{m}{k} } \\T_2=2\pi \sqrt{\frac{m+2}{k} } \\\frac{T_1}{T_2} =\sqrt{\frac{m}{m+2} }\\\frac{2}{3} =\sqrt{\frac{m}{m+2} }\\\\

               m=\frac{5}{8} =0.625kg

Thus, we can conclude that, the mass m will be 0.625kg.

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