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3 years ago

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A uniform rod of length L is pivoted at L/4 from one end. It is pulled to one side through a very small angle and allowed to osc

illate in a vertical plane. Determine the period of oscillation of the pendulum if L = 5.95 m.

1 answer:

ludmilkaskok [199]3 years ago

8 0

Answer:

T= 4.24sec

Explanation:

We are going to use the formula below to calculate.

$T=2\pi \sqrt{\frac{L}{g} }$

Where T is period

L is length of rod

g is acceleration due to gravity = $9.8m/s^{2}$

From the problem, the rod is pivoted at 1/4L which means that three quarter of the rod was used for the oscillation. lets call this $L_{O}$

$L_{O} = 3/4 * 5.95m$

= 4.4625m

thus $T=2\pi \sqrt{\frac{L_{O} }{g} }$

$T=2\pi \sqrt{\frac{4.4625 }{9.8} }$

T= 4.24sec

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Answer:

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The proton has a smaller wavelength than the electron.

$\lambda_{proton} = 6.05x10^{-14}m$ < $\lambda_{electron} = 1.10x10^{-10}m$

Part B:

The proton has a smaller wavelength than the electron.

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The wavelength of each particle can be determined by means of the De Broglie equation.

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Case for the electron:

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But $J = Kg.m^{2}/s^{2}$

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(9.11x10^{-31}Kg)(6.55x10^{6}m/s)}$

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Case for the proton:

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First, it is necessary to find the velocity associated to that kinetic energy:

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$v = \sqrt{\frac{2(7.89x10^{-15}kg \cdot m^{2}/s^{2})}{9.11x10^{-31}Kg}}$

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Then, equation 2 can be used:

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$\lambda = 1.29x10^{-13}m$

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