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3 years ago

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A damped oscillator is released from rest with an initial displacement of 10.00 cm. At the end of the first complete oscillation

, the displacement reaches 9.05 cm. When 4 more oscillations are completed, what is the displacement reached

1 answer:

Tpy6a [65]3 years ago

5 0

Answer:

The displacement is $A_r = 6.071 \ cm$

Explanation:

From the question we are told that

The initial displacement is $A_o = 10 \ cm$

The displacement at the end of first oscillation is $A_d = 9.05 \ cm$

Generally the damping constant of this damped oscillator is mathematically represented as

$\eta = \frac{A_d}{A_o}$

substituting values

$\eta = \frac{9.05}{10}$

$\eta = 0.905$

The displacement after 4 more oscillation is mathematically represented as

$A_r = \eta^4 * A_d$

substituting values

$A_r = (0.905)^4 * (9.05)$

$A_r = 6.071 \ cm$

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Explanation:

Given:

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Electric field on the axis of the ring of radius 'r' at a distance of 'x' from the center of the ring is given as:

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Plug in the given values for each point and solve.

(a)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=1.00\ cm=0.01\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

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(b)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=5.00\ cm=0.05\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.05)}{((0.05)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{33750}{1.3975\times 10^{-3}}\\\\E_x=24150268.34\ N/C$

(c)

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$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=30.0\ cm=0.30\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.30)}{((0.30)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{202500}{0.0316}\\\\E_x=6408227.848\ N/C$

(d)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=100\ cm=1\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(1)}{((1)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{675000}{1.015}\\\\E_x=665024.6305\ N/C$

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3 years ago