Vật lý 12

**phannguyenquoctu** Thu May 28, 2015 11:58 am

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**phannguyenquoctu** Thu May 28, 2015 11:58 am

**phannguyenquoctu** Thu May 28, 2015 12:31 pm

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**phannguyenquoctu** Thu May 28, 2015 12:41 pm

Phase Difference and Phase Shift

Phase Difference

In the last tutorial, we saw that the [Only admins are allowed to see this link] (Sine Wave) can be presented graphically in the time domain along an horizontal zero axis, and that sine waves have a positive maximum value at time π/2, a negative maximum value at time 3π/2, with zero values occurring along the baseline at 0, π and 2π. However, not all sinusoidal waveforms will pass exactly through the zero axis point at the same time, but may be “shifted” to the right or to the left of 0o by some value when compared to another sine wave.

For example, comparing a voltage waveform to that of a current waveform. This then produces an angular shift or Phase Difference between the two sinusoidal waveforms. Any sine wave that does not pass through zero at t = 0 has a phase shift.

The phase difference or phase shift as it is also called of a [Only admins are allowed to see this link] is the angle Φ(Greek letter Phi), in degrees or radians that the waveform has shifted from a certain reference point along the horizontal zero axis. In other words phase shift is the lateral difference between two or more waveforms along a common axis and sinusoidal waveforms of the same frequency can have a phase difference.

The phase difference, Φ of an alternating waveform can vary from between 0 to its maximum time period, T of the waveform during one complete cycle and this can be anywhere along the horizontal axis between, Φ = 0 to 2π (radians) or Φ = 0 to 360o depending upon the angular units used.

Phase difference can also be expressed as a time shift of τ in seconds representing a fraction of the time period, T for example, +10mS or – 50uS but generally it is more common to express phase difference as an angular measurement.

Then the equation for the instantaneous value of a sinusoidal voltage or current waveform we developed in the previous [Only admins are allowed to see this link] will need to be modified to take account of the phase angle of the waveform and this new general expression becomes.

Phase Difference Equation

[Only admins are allowed to see this image]

Where:
Am - is the amplitude of the waveform.
ωt - is the angular frequency of the waveform in radian/sec.
Φ (phi) - is the phase angle in degrees or radians that the waveform has shifted either left or right from the reference point.

If the positive slope of the sinusoidal waveform passes through the horizontal axis “before” t = 0then the waveform has shifted to the left so Φ >0, and the phase angle will be positive in nature, +Φgiving a leading phase angle. In other words it appears earlier in time than 0o producing an anticlockwise rotation of the vector.

Likewise, if the positive slope of the sinusoidal waveform passes through the horizontal x-axis some time “after” t = 0 then the waveform has shifted to the right so Φ <0, and the phase angle will be negative in nature -Φ producing a lagging phase angle as it appears later in time than 0o producing a clockwise rotation of the vector. Both cases are shown below.

Phase Relationship of a Sinusoidal Waveform

[Only admins are allowed to see this image]

Firstly, lets consider that two alternating quantities such as a voltage, v and a current, i have the same frequency ƒ in Hertz. As the frequency of the two quantities is the same the angular velocity,ω must also be the same. So at any instant in time we can say that the phase of voltage, v will be the same as the phase of the current, i.

Then the angle of rotation within a particular time period will always be the same and the phase difference between the two quantities of v and i will therefore be zero and Φ = 0. As the frequency of the voltage, v and the current, i are the same they must both reach their maximum positive, negative and zero values during one complete cycle at the same time (although their amplitudes may be different). Then the two alternating quantities, v and i are said to be “in-phase”.

Two Sinusoidal Waveforms – “in-phase”

[Only admins are allowed to see this image]

Now lets consider that the voltage, v and the current, i have a phase difference between themselves of 30o, so (Φ = 30o or π/6 radians). As both alternating quantities rotate at the same speed, i.e. they have the same frequency, this phase difference will remain constant for all instants in time, then the phase difference of 30o between the two quantities is represented by phi, Φ as shown below.

Phase Difference of a Sinusoidal Waveform

[Only admins are allowed to see this image]

The voltage waveform above starts at zero along the horizontal reference axis, but at that same instant of time the current waveform is still negative in value and does not cross this reference axis until 30o later. Then there exists a Phase difference between the two waveforms as the current cross the horizontal reference axis reaching its maximum peak and zero values after the voltage waveform.

As the two waveforms are no longer “in-phase”, they must therefore be “out-of-phase” by an amount determined by phi, Φ and in our example this is 30o. So we can say that the two waveforms are now 30o out-of phase. The current waveform can also be said to be “lagging” behind the voltage waveform by the phase angle, Φ. Then in our example above the two waveforms have a Lagging Phase Difference so the expression for both the voltage and current above will be given as.

[Only admins are allowed to see this image]

where, i lags v by angle Φ

Likewise, if the current, i has a positive value and crosses the reference axis reaching its maximum peak and zero values at some time before the voltage, v then the current waveform will be “leading” the voltage by some phase angle. Then the two waveforms are said to have a Leading Phase Difference and the expression for both the voltage and the current will be.

[Only admins are allowed to see this image]

where, i leads v by angle Φ

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The phase angle of a sine wave can be used to describe the relationship of one sine wave to another by using the terms “Leading” and “Lagging” to indicate the relationship between two sinusoidal waveforms of the same frequency, plotted onto the same reference axis. In our example above the two waveforms are out-of-phase by30o so we can say that i lags v or v leads i by 30o.

The relationship between the two waveforms and the resulting phase angle can be measured anywhere along the horizontal zero axis through which each waveform passes with the “same slope” direction either positive or negative.

In AC power circuits this ability to describe the relationship between a voltage and a current sine wave within the same circuit is very important and forms the bases of AC circuit analysis.

The Cosine Waveform

So we now know that if a waveform is “shifted” to the right or left of 0o when compared to another sine wave the expression for this waveform becomes Am sin(ωt ± Φ). But if the waveform crosses the horizontal zero axis with a positive going slope 90o or π/2 radians before the reference waveform, the waveform is called a Cosine Waveform and the expression becomes.

Cosine Expression

[Only admins are allowed to see this image]

The Cosine Wave, simply called “cos”, is as important as the sine wave in electrical engineering. The cosine wave has the same shape as its sine wave counterpart that is it is a sinusoidal function, but is shifted by +90o or one full quarter of a period ahead of it.

Phase Difference between a Sine wave and a Cosine wave

[Only admins are allowed to see this image]

Alternatively, we can also say that a sine wave is a cosine wave that has been shifted in the other direction by -90o. Either way when dealing with sine waves or cosine waves with an angle the following rules will always apply.

Sine and Cosine Wave Relationships

[Only admins are allowed to see this image]

When comparing two sinusoidal waveforms it more common to express their relationship as either a sine or cosine with positive going amplitudes and this is achieved using the following mathematical identities.

[Only admins are allowed to see this image]

By using these relationships above we can convert any sinusoidal waveform with or without an angular or phase difference from either a sine wave into a cosine wave or vice versa.

In the next tutorial about [Only admins are allowed to see this link] we will use a graphical method of representing or comparing the phase difference between two sinusoids by looking at the phasor representation of a single phase AC quantity along with some phasor algebra relating to the mathematical addition of two or more phasors.

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**phannguyenquoctu** Thu May 28, 2015 12:50 pm

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**phannguyenquoctu** Thu May 28, 2015 1:15 pm

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**phannguyenquoctu** Thu May 28, 2015 1:16 pm

**phannguyenquoctu** Sat May 30, 2015 11:29 am

Giải phương trình bằng lệnh SOLVE trên máy tính Casio fx570ES PLUS
Thứ bảy, 05/01/2013, 03:03 GMT+7

Tìm nghiệm ( Radian) của phương trình: [Only admins are allowed to see this image] [Only admins are allowed to see this image]

Bài làm
Trên máy tính Casio fx570ES PLUS
Chọn màn hình Radian và ghi vào màn hình [Only admins are allowed to see this image]
Ấn SHIFT SOLVE và chọn giá trị đầu là 0,5 chẳng hạn ấn = ta được kết quả [Only admins are allowed to see this image] [Only admins are allowed to see this image]
Ghi chú chọn giá trị đầu nào khác nữa trong [Only admins are allowed to see this image] vẫn được nghiệm duy nhất là [Only admins are allowed to see this image]
( bài này nếu giải bằng phương pháp thông thường thì rất rắc rối ! )

Người viết : sangnguyen

**phannguyenquoctu** Sat May 30, 2015 11:34 am

using SOLVE

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**phannguyenquoctu** Sat May 30, 2015 11:35 am

Giải phương trình bằng lệnh Solve trên máy tính Casio fx570ES PLUS
Thứ năm, 01/11/2012, 13:00 GMT+7

Tìm x biết [Only admins are allowed to see this image]

Trích đề thi giỏi toán trên máy tính Casio 26/06/2012

Giải trên máy tính Casio fx570ES PLUS
Ta ghi vào màn hình
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Rồi ấn SHIFT Solve rồi cho x= 1,2,3... chẳng hạng rồi ấn =
Ta được kết quả 13,335 9264
Vậy x=13,335 9264

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Phase shift

Example 1

Sketch the curve
y = sin(2x + 1)

Let's first graph y=sin 2x (without the "+ 1") to get an idea of what phase shift means.
The [Only admins are allowed to see this link] is 1 and the [Only admins are allowed to see this link] is

2

2π

=π.
I have drawn a little more than 2 cycles, starting from x=0.

[Only admins are allowed to see this image]

Now let's consider the phase shift. Using the formula above, we will need to shift our curve by:

Phase shift =−

b

c
=−

2

1
=−0.5

This means we have to shift the curve to the left (because the phase shift is negative) by 0.5. Here is the answer (in blue). I have kept the original y = sin 2x (in dotted gray) so you can see what's happening.

[Only admins are allowed to see this image]

Note that the curve passes through:

(-0.5, 0) on the x-axis (because we shifted the curve to the left by 0.5) and
sin 1 = 0.84 on the y-axis. This figure is obtained by substituting x = 0 into y = sin(2x + 1); (in radians, of course).

Example 2

Sketch

y=12cos(2x−

8

π
)

In this example, the

amplitude a=12, and

b=2, so the period =

b

2π
=

2

2π
=π, and

c=−

8

π

so the phase shift is

−

b

c
=−

2

−π/8
=

16

π

This means we need to move the cosine graph to the right of its normal position (because the displacement is positive in this example) by

16

π

.
Without phase shift, the cosine curve will be y = 12 cos 2x and its graph is as follows:

[Only admins are allowed to see this image]

Now, let's shift it to the right by π/16 ≈ 0.1963:

[Only admins are allowed to see this image]

You can see the original cosine curve (in gray) and we have shifted it by

16

π

to the right to give the required graph of y=12cos(2x−

8

π

), in blue.

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**phannguyenquoctu** Sat May 30, 2015 11:38 am

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Dùng máy tính Casio fx-570ES và 570ES Plus để giải nhanh một số bài tập trắc nghiệm vật lý 12

**phannguyenquoctu** Sat May 30, 2015 11:43 am

Chức năng SOLVE trên máy tính Casio FX-570MS
Thứ năm, 02/08/2012, 09:13 GMT+7

Chức năng SOLVE trên máy tính Casio FX-570MS giúp ta giải một biểu thức theo các giá trị biến khác mà không cần biến đổi hay đơn giản.

Ví dụ 1: C là thời gian để 1 vật có vận tốc ban đầu A đạt độ cao B.

Dùng công thức sau để tính A khi B=14m với C=2 giây và gia tốc D=9,8m/s2.

Kết quả A=16.8 B=AC - 12DC2

[Only admins are allowed to see this image]
Phương trình chứa căn thức
Một số phương trình chứa ẩn trong dấu căn bậc hai cũng có thể tìm được nghiệm (gần đúng) bằng lệnh SOLVE
“Theo sách hướng dẫn sử dụng máy tính FX-570MS”

**phannguyenquoctu** Sun May 31, 2015 5:23 pm

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Chiều dài của con lắc lò xo khi treo thẳng đứng

1. Phương pháp
Tại vị trí cân bằng: ℓcb = ℓ0 + Δℓ0
Tại vị trí li độ x: ℓ = ℓcb + x
A ≤ Δℓ0 → khi dao động lò xo luôn bị giãn.

Giãn ít nhất ( khi vật ở vị trí cao nhất) Δℓ0 – A.
Giãn nhiều nhất ( khi vật ở vị trí thấp nhất) Δℓ0 + A.

A > Δℓ0 → khi dao động lò xo vừa giãn vừa nén.

Nén nhiều nhất khi A - Δℓ0.
Không biến dạng khi x = - Δℓ0.
Giãn nhiều nhất ( khi vật ở vị trí thấp nhất) A + Δℓ0.

2. Vận dụng

Ví dụ 1:
Một vật treo vào lò xo làm nó giãn ra 4cm. Cho g =π2=10m/s2. Biết lực đàn hồi cực đại và cực tiểu lần lượt là 10N và 6N. Chiều dài tự nhiên của lò xo 20cm. Chiều dài cực tiểu và cực đại của lò xo trong quá trình dao động là
A. 25cm và 24cm.
B. 24cm và 23cm.
C. 26cm và 24cm.
D. 25cm và 23cm.

Lời giải

Lực đàn hồi cực đại: Fmax = k(Δℓ0 + A)
Lực đàn hồi cực tiểu có giá trị lớn hơn 0 nên Δℓ0 > A: Fmin = k(Δℓ0 – A)
Ta có tỉ số: [ltr][size=12]FmaxFmin=k(Δl0+A)k(Δl0−A)=Δl0+AΔl0−A→106=0,04+A0,04−A→0,4−10A=0,24+6A→A=0,4−0,2416=0,01m[/ltr][/size]
Chiều dài cực đại ℓmax = ℓ + Δℓ0 + A = 20 + 4 + 1 = 25cm
Chiều dài cực tiểu ℓmin = ℓ + Δℓ0 - A = 20 + 4 - 1 = 23cm

Chọn D
Ví dụ 2:
Một con lắc lò xo treo thẳng đứng, đầu trên cố định, đầu dưới treo một vật m = 100g. Kéo vật xuống dưới vị trí cân bằng theo phương thẳng đứng rồi buông nhẹ. Vật dao động theo phương trình x = 5cos(4πt + π/2)cm. Chọn gốc thời gian là lúc buông vật, lấy g = 10m/s2. Lực dùng để kéo vật trước khi dao động có độ lớn
A. 1,8N
B. 6,4N
C. 0,8N
D. 3,2N

Lời giải

Khi kéo vật tới li độ x0 rồi buông nhẹ, vậy x = x0 = A
Nghĩa là vật ở biên độ, vậy lực cần kéo đó tương ứng với lực đàn hồi cực đại tác dụng vào con lắc lò xo: Fmax = k(Δℓ0 + A)
Mặt khác: [ltr]ω[size=12]2=km=gΔl0→⎡⎣⎢Δl0=gω2=10(4π)2=0,0625mk=gmΔl0=0,1.100,0625=16N/m[/ltr][/size]
Thế Δℓ0 vào biểu thức trên ta có: Fmax = 16(0,0625 + 0,05) = 1,8 N

Chọn A

Ví dụ 3:
Một con lắc lò xo treo thẳng đứng, vật treo có khối lượng m. Kéo vật xuống dưới vị trí cân bằng 3 cm rồi truyền cho nó vận tốc 40 cm/s thì nó dao động điều hòa theo phương thẳng đứng trùng với trục của lò xo và khi vật đạt độ cao cực đại, lò xo giãn 5 cm. Lấy gia tốc trọng trường g = 10 m/s2. Vận tốc cực đại của vật dao động là
A. 1,15 m/s.
B. 0,5 m/s.
C. 10 cm/s.
D. 2,5 cm/s.

Lời giải

[ltr]Δℓ[size=12]min=Δℓ0−A→Δℓmin=gω2−A→0,05=10ω2−A→1ω2=0,005+0,1AA2=x20+v20ω2→A2=x20+v20ω2=0,032+0,42.(0,005+0,1A)→A=0,05(m)→ω=10(rad/s)→vmax=0,5(m/s)[/ltr]
Chọn B

Bài tập về nhà
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Phiếu đáp án: [Only admins are allowed to see this link][/size]

**phannguyenquoctu** Sun May 31, 2015 5:30 pm

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**phannguyenquoctu** Mon Jun 01, 2015 10:21 am

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Vật lý 12

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Phase Difference

Phase Difference Equation

Phase Relationship of a Sinusoidal Waveform

Two Sinusoidal Waveforms – “in-phase”

Phase Difference of a Sinusoidal Waveform

The Cosine Waveform

Cosine Expression

Phase Difference between a Sine wave and a Cosine wave

Sine and Cosine Wave Relationships

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Example 1

Example 2

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