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2 years ago

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Harmonics problem. A square wave of frequency f contains harmonics (sine waves) at f, 3f, 5f, 7f, ... . Suppose a system respond

s to frequencies in the range 20-40 kHz but is insensitive outside of this range.
EXAMPLE:
Imagine applying a square wave with f = 30 kHz to this system. What is the shape of the response?
[Answer: a 30 kHz sine wave. ]
Imagine applying a square wave with f = 10 kHz to this system.

What is the shape of the response?
A. A 10 kHz sine wave
B. A combination of 20, 30 and 40 kHz sine waves
C. A 30 kHz sine wave
D. No response

1 answer:

ad-work [718]2 years ago

4 0

Answer:

C.

Explanation:

If the fundamental frequency of the square wave is 10 Khz, and contains harmonics at f, 3f, 5f, etc., it will have a component at 10 Khz, other at 30 Khz, and another one at 50 Khz.
If the system responds to frequencies in the range between 20 Khz and 40 KHz only, the shape of the response will contain only a 30 Khz sine wave (3rd harmonic of 10 Khz).

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I attached the missing picture.
We can figure this one out using the law of conservation of energy.
At point A the car would have potential energy and kinetic energy.
$A: mgh_1+\frac{mv_1^2}{2}$
Then, while the car is traveling down the track it loses some of its initial energy due to friction:
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So, we know that the car is approaching the point B with the following amount of energy:
$mgh_1+\frac{mv_1^2}{2}- F_fL$
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2 years ago

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wheel rotates from rest with constant angular acceleration. Part A If it rotates through 8.00 revolutions in the first 2.50 s, h

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Answer:

x2 = 64 revolutions.

it rotate through 64 revolutions in the next 5.00 s

Explanation:

Given;

wheel rotates from rest with constant angular acceleration.

Initial angular speed v = 0

Time t = 2.50

Distance x = 8 rev

Applying equation of motion;

x = vt +0.5at^2 ........1

Since v = 0

x = 0.5at^2

making a the subject of formula;

a = x/0.5t^2 = 2x/t^2

a = angular acceleration

t = time taken

x = angular distance

Substituting the values;

a = 2(8)/2.5^2

a = 2.56 rev/s^2

velocity at t = 2.50

v1 = a×t = 2.56×2.50 = 6.4 rev/s

Through the next 5 second;

t2 = 5 seconds

a2 = 2.56 rev/s^2

v2 = 6.4 rev/s

From equation 1;

x = vt +0.5at^2

Substituting the values;

x2 = 6.4(5) + 0.5×2.56×5^2

x2 = 64 revolutions.

it rotate through 64 revolutions in the next 5.00 s

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2 years ago

A small crack occurs at the base of a 15.0-m-high dam. The effective area through which water leaves is 2.30 × 10-3 m2. (a) Igno

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Answer

given,

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effective area of water = 2.3 x 10⁻³ m²

Using energy conservation

$m g h = \dfrac{1}{2}mv^2$

$v= \sqrt{2gh}$

$v= \sqrt{2\times 9.8 \times 15}$

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discharge of water

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2 years ago

A guitar string has a linear density of 8.30 ✕ 10−4 kg/m and a length of 0.660 m. the tension in the string is 56.7 n. when the

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Ans: Beat Frequency = 1.97Hz

Explanation:
The fundamental frequency on a vibrating string is

$f = \sqrt{ \frac{T}{4mL} }$ <span> -- (A)</span>

<span>here, T=Tension in the string=56.7N,
L=Length of the string=0.66m,
m= mass = 8.3x10^-4kg/m * 0.66m = 5.48x10^-4kg </span>

Plug in the values in Equation (A)

<span>so </span> $f = \sqrt{ \frac{56.7}{4*5.48*10^{-4}*0.66} }$ <span> = 197.97Hz </span>

<span>the beat frequency is the difference between these two frequencies, therefore:
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2 years ago

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Answer

given,

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so, from the diagram attached below

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$F_b sin {\theta_b} = 2.989 -1.303$

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$F_b cos {\theta_b} = F_L sin {\theta_L}$

$cos {\theta_b} = \dfrac{F_L sin {\theta_L}}{F_b}$

$cos {\theta_b} = \dfrac{1.52 \times sin {31^0}}{1.686}$

$cos {\theta_b} = 0.464$

θ = 62.35°

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