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2 years ago

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An accelerometer has a damping ratio of 0.5 and a natural frequency of 18,000 Hz. It is used to sense the relative displacement

of a beam to which it is attached. (a)If an impact to the beam imparts a vibration at 4500 Hz, calculate the dynamic error and phase shift in the accelerometer output. (b)Calculateits resonance frequency.(c)What isthe maximumpossiblemagnitude ratio that the system can achieve

1 answer:

SOVA2 [1]2 years ago

3 0

Answer:

A) i) Dynamic error ≈ 3.1%

ii) phase shift ≈ -12°

B) 79971.89 rad/s

Explanation:

Given data :

Damping ratio = 0.5

natural frequency = 18,000 Hz

<u>a) Calculate the dynamic error and phase shift in accelerometer output at an impart vibration of 4500 Hz</u>

i) Dynamic error

This can be calculated using magnitude ratio formula attached below is the solution

dynamic error ≈ 3.1%

ii) phase shift

This phase shift can be calculated using frequency dependent phase shift formula

phase shift ≈ -12°

<u>B) Determine resonance frequency </u>

Wr = 2 $\pi$ ( 18000 $\sqrt{0.5}$ ) = 79971.89 rad/s

C) The maximum magnitude ratio that the system can achieve

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A power supply has an open-circuit voltage of 40.0 V and an internal resistance of 2.00 $\Omega$ . It is used to charge two storage batteries connected in series, each having an emf of 6.00 V and internal resistance of 0.300 $\Omega$ . If the charging current is to be 4.00 A, (a) what additional resistance should be added in series? At what rate does the internal energy increase in (b) the supply, (c) in the batteries, and (d) in the added series resistance? (e) At what rate does the chemical energy increase in the batteries?

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Explanation:

From the question we are told that

The open circuit voltage is $V = 40.0V$

The internal resistance is $R = 2 \Omega$

The emf of each battery is $e = 6.00 V$

The internal resistance of the battery is $r = 0.300V$

The charging current is $I = 4.00 \ A$

Let assume the the additional resistance to to added to the circuit is $R_z$

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The total resistance in the circuit is

$R_T = R + 2r +R_z$

Substituting values

$R_T = 2.6 +R_z$

And the difference in potential in the circuit is

$E = V -2e$

=> $E = 40 - (2 * 6)$

$E = 28 V$

Now according to ohm's law

$I = \frac{E}{R_T}$

Substituting values

$4 = \frac{28}{R_z + 2.6}$

Making $R_z$ the subject of the formula

So $R_z = \frac{28 - 10.4}{4}$

$R_z = 4.4 \Omega$

The increase rate of internal energy at the supply is mathematically represented as

$Z_1 = I^2 R$

Substituting values

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$Z_2 = I^2 r$

Substituting values

$Z_2 = 4^2 * 2 * 0.3$

$Z_2 = 9.6 \ W$

The increase rate of internal energy at the added series resistance is mathematically represented as

$Z_3 = I^2 R_z$

Substituting values

$Z_3 = 4^2 * 4.4$

$Z_3 = 70.4 W$

Generally the increase rate of the chemically energy in the battery is mathematically represented as

$C = 2 * e * I$

Substituting values

$C = 2 * 6 * 4$

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