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sin θ = sin π θ

sin π 3 = sin π - π 3 = sin 2 π 3 = 1 2

We can determine the slope of the waveform by partially differentiating y-function with respect to "x" (considering "t" constant) :

y t = t A sin k x ω t + φ = k A cos k x w t + φ

At x = 0, t = 0, φ = π / 3

Slope = k A cos φ = k A cos π 3 = k A X 3 2 = a positive number

At x = 0, t = 0, φ = 2 π / 3

Slope = k A cos φ = k A cos 2 π 3 = k A X 3 2 = a negative number

The smaller of the two (π/3) in first quadrant indicates that wave form has positive slope and is increasing as we move along x-axis. The greater of the two (2π/3) similarly indicates that wave form has negative slope and is decreasing as we move along x-axis. The two initial wave forms corresponding to two initial phase angles in first and second quadrants are shown in the figure below.

Initial phase

Initial phase angles in first and second quadrant.

We can interpret initial phase angle in the third and fourth quadrants in the same fashion. The sine values of angles in third and fourth quadrants are negative. There is a pair of two angles for which sine has equal negative values. The angle in third quarter like 4π/3 indicates that wave form has negative slope and is further decreasing (more negative) as we move along x-axis. On the other hand, corresponding angle in fourth quadrant for which magnitude is same is 5π/3.

sin 4 π 3 = sin 5 π 3 = - 1 2

For initial phase angle of 5π/3 in fourth quadrant, the wave form has positive slope and is increasing (less negative) as we move along x-axis. The two initial wave forms corresponding to two initial phase angles in third and fourth quadrants are shown in the figure below.

Initial phase

Initial phase angles in third and fourth quadrant.

We can also denote initial phase angles in third and fourth quadrants (angles greater than “π”) as negative angles, measured clockwise from the reference direction. The equivalent negative angles for the example here are :

2 π 4 π 3 = 4 π 3 2 π = 2 π 3

and

2 π 5 π 3 = 5 π 3 2 π = π 3

Particle velocity and acceleration

Particle velocity at a given position x=x is obtained by differentiating wave function with respect to time “t”. We need to differentiate equation by treating “x” as constant. The partial differentiation yields particle velocity as :

v p = t y x , t = t A sin k x ω t = ω A cos k x ω t

We can use the property of cosine function to find the maximum velocity. We obtain maximum speed when cosine function evaluates to “-1” :

v p max = ω A

The acceleration of the particle is obtained by differentiating expression of velocity partially with respect to time :

a p = t v p = t { ω A cos k x ω t } = - ω 2 A sin k x ω t = - ω 2 y

Again the maximum value of the acceleration can be obtained using property of sine function :

a p max = ω 2 A

Relation between particle velocity and wave (phase) speed

We have seen that particle velocity at position “x” and time “t” is obtained by differentiating wave equation with respect to “t”, while keeping time “x” constant :

v p = t y x , t = ω A cos k x ω t

On the other hand, differentiating wave equation with respect to “x”, while keeping “t”, we have :

x y x , t = k A cos k x ω t

The partial differentiation gives the slope of wave form. Knowing that speed of the wave is equal to ratio of angular frequency and wave number, we divide first equation by second :

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
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cm
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A mouse of mass 200 g falls 100 m down a vertical mine shaft and lands at the bottom with a speed of 8.0 m/s. During its fall, how much work is done on the mouse by air resistance
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Can you compute that for me. Ty
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Chemistry is a branch of science that deals with the study of matter,it composition,it structure and the changes it undergoes
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A ball is thrown straight up.it passes a 2.0m high window 7.50 m off the ground on it path up and takes 1.30 s to go past the window.what was the ball initial velocity
Krampah Reply
2. A sled plus passenger with total mass 50 kg is pulled 20 m across the snow (0.20) at constant velocity by a force directed 25° above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.
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you have been hired as an espert witness in a court case involving an automobile accident. the accident involved car A of mass 1500kg which crashed into stationary car B of mass 1100kg. the driver of car A applied his brakes 15 m before he skidded and crashed into car B. after the collision, car A s
Samuel Reply
can someone explain to me, an ignorant high school student, why the trend of the graph doesn't follow the fact that the higher frequency a sound wave is, the more power it is, hence, making me think the phons output would follow this general trend?
Joseph Reply
Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
Joseph
Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
Joseph
"Generation of electrical energy from sound energy | IEEE Conference Publication | IEEE Xplore" ***ieeexplore.ieee.org/document/7150687?reload=true
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progressive wave
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A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
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Source:  OpenStax, Oscillation and wave motion. OpenStax CNX. Apr 19, 2008 Download for free at http://cnx.org/content/col10493/1.12
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