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3 years ago

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A simple pendulum with a length of 2.23 m and a mass of 6.74 kg is given an initial speed of 2.06 m/s at its equi- librium posit

ion. Assume it undergoes simple harmonic motion. Determine (a) its period, (b) its total energy, and (c) its maximum angular displacement.

1 answer:

solniwko [45]3 years ago

4 0

Answer:

a) T = 2.997 s

b) K = 14.3 J

c) φ = 0.444 rad

Explanation:

a) Determine its period

The pendulum simple’s period is:

T = 2π $\sqrt{\frac{l}{g} }$

Where l: Pendulum’s length

g = 9.8 m/s2

T = 2π $\sqrt{\frac{2.23}{9.8} }$

T = 2.997 s

b) Total energy

Initially his total energy is kinetic

K = $\frac{mv^{2} }{2}$

K = $\frac{(6.74)(2.06)^{2} }{2}$

K = 14.3 J

c) Maximum angular displacement

φ = $cos^{-1}(1-\frac{E}{mgl} )$

φ = $cos^{-1}(1-\frac{14.3}{(6.74)(9.8)(2.23)} )$

φ = 0.444 rad

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$\frac{d}{da} \kappa = \frac{1}{a^2 + b^2} \frac{d}{da}(a)+ a * \frac{d}{da} (\frac{1}{a^2 + b^2} )$

$\frac{d}{da} \kappa = \frac{1}{a^2 + b^2} * 1 + a * (-1) (\frac{1}{(a^2 + b^2)^2} ) \frac{d}{da} (a^2+b^2)$

$\frac{d}{da} \kappa = \frac{1}{a^2 + b^2} * 1 - a (\frac{1}{(a^2 + b^2)^2} ) (2* a)$

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We dcan skip solving the equation noting that, if a=b, then

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at this point, this give us only the first term

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if a is greater than zero, this means that the second derivative is negative, and the point is a minimum

the value of kappa is

$\kappa = \frac{b}{b^2 + b^2}$

$\kappa = \frac{b}{2* b^2}$

$\kappa = \frac{1}{2 b}$

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