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In this Physics tutorial, you will learn:

- What is wave function?
- What is the form of the general equation of waves?
- What is angular wave number and how it relates with the wavelength?
- How to obtain the simplified wave equation from the general equation of waves?
- How to calculate the speed and acceleration of oscillating particles in a wave?
- How to interpret wave graphs?

In the previous tutorial "Types of Waves. The Simplified Equation of Waves", we explained some key concepts that are related to waves and their behavior. Also, in that tutorial we gave the simplified equation of waves

v = λ × f

However, there is a serious drawback in the above equation. It does not include the amplitude, which is a key element of a wave. Thus, if we apply the above equation, we are able to use only the information contained in the horizontal axis. Every quantity related to the vertical or y-axis such as amplitude, crest, trough and any other vertical position of a point of wave, cannot be calculated if they are not shown a-priori, and as a result, it is impossible to find them.

In this tutorial, besides speed, frequency, period, wavelength etc., we will extend the use of equation of waves for the vertical axis as well, in order to include all relevant quantities, so that we can find the position of a point of a point of wave at any instant.

As you may have noticed in the previous tutorial, the waves motion is periodical. This means it has many similarities with both circular motion and SHM. The graph of waves is sinusoidal just like in SHM.

However, as stated above (and as seen in the examples of the previous tutorial), since we are dealing with travelling waves, i.e. with waves that are displaced from the original position, when studying their behavior, we need to consider three quantities instead of two: the x- and y-coordinates versus time t. Therefore, the equation of waves must be of the type

y(x,t)

where the horizontal position x and the time t are taken as independent variables (values are inserted by the user) while the vertical position y is to be calculated (it is a dependent variable).

Remember that in none of the other abovementioned topics we had such a situation involving three variables. Two variables (one independent and the other dependent) were enough to describe in full such motions. Thus, in circular motion, the equation has the form r(t) where r is the radius of circle and t is the time, while in SHM, the equation of motion is of the form x(t) or θ(t) depending on the type of SHM (linear or rotational).

Hence, given that the wave equation has a sinusoidal form, we write its general form as

y(x,t) = y_{max} × sin(k × x - ω × t)

You are familiar with most quantities shown in the above equation. Thus,

y_{max} represents the amplitude A of a point of the wave, as it oscillates according the y-axis,

ω represents the angular frequency (ω = 2π / T as usual),

x is the horizontal position,

t is the time of motion, and

y(x, t) is the vertical position at a given instant.

The quantity k is known as the **angular wave number**. Let's find its relationship with the other known quantities discussed above. Thus, at t = 0, we have

y(x_{0},0) = y_{max} × sin(k × x_{0} )

When one cycle is completed, the y-values (the displacement) repeat themselves while the x-values increase by one wavelength λ. Thus, at x = x1 we have

x_{1} = x_{0} + λ

As stated above, we have the same y-coordinate (y_{0} = y_{1}) for two x-coordinates that change by λ units after a time t1 = T (period). Hence, the equation of wave becomes

y(x_{1},T) = y_{max} × sin(k × x_{1} )

= y_{max} × sin[k × (x_{0} + λ)]

= sin(k × x_{0} + k × λ)

= y

= sin(k × x

Given that

y(x_{0},0) = y(x_{1},T)

and since the sine functions repeat themselves after every 2π change in x-coordinate, we obtain

k × λ = 2π

Therefore, we obtain for the angular wave number k

k = *2π**/**λ*

In this way, the wave equation becomes

y(x,t) = y_{max} × sin(*2π**/**λ* × x - ω × t)

Also, since ω = ** 2π/T**, we can also insert the period T in the wave equation. Therefore, it becomes

y(x,t) = y_{max} × sin(*2π**/**λ* × x - *2π**/**T* × t)

In this way, we inserted the values of wavelength and period in the equation of waves, so now it is more complete.

Look at the figure.

It must be noted that the argument

k × x - ω × t

or

represents the phase of the wave equation.

If there is a phase shift φ from the original position of wave, we insert its value in the equation of waves. Thus, it becomes

y(x,t) = y_{max} × sin(k × x - ω × t + φ)

**Remark!** The sign minus in the argument (k × x - ω × t) is when the wave moves in the positive direction. When the wave moves towards negative, the argument takes the positive sign, i.e. it becomes (k × x + ω × t).

Calculate the y-coordinate of the wave shown in the figure below at t = 2.4 s using the information provided in the graph.

From the graph we see that amplitude y_{max} = 8 m. Also, the number N of cycles when the wave reaches the point A = 3. Given that the wave in that instant is at x_{A} = 30 m, we obtain for the wavelength

λ = *x*_{A}*/**N* = *30 m**/**3* = 10 m

The wave starts from the origin, therefore φ = 0.

Also, given that t_{A} = 24 s, we obtain for the period T:

T = *t*_{A}*/**N* = *24 s**/**3* = 8s

The horizontal position x at t = 2.4 s can be calculated through the cross-product method. We can denote by B the point at which the wave is at t_{B} = 2.4 s. Thus, given that the wave moves at constant speed, we can write

(x_{A} - x_{0})/(t_{A} - t_{0} ) = (x_{B} - x_{0})/(t_{B} - t_{0} )

*30 m - 0**/**24 s - 0* = *x*_{B} - 0*/**2.4 s - 0*

*30**/**24* = *x*_{B}*/**2.4*

x_{B} = *30 × 2.4**/**24*

= 3 m

x

= 3 m

Now, we have everything we need to calculate the vertical position at the given instant. Thus, from the equation

y(x,t) = y_{max} × sin(*2π**/**λ* × x - *2π**/**T* × t)

we find after substituting the values,

y(3 m,2.4 s) = 8 × sin(*2π**/**10* × 3 - *2π**/**8* × 2.4)

= 8 × sin(*6π**/**10* - *4.8π**/**8*)

= 8 × sin(

Given that both fractions give the same value, we can find the y-value by ignoring one independent variable, for example t. Thus, we obtain

y(3m) = 8 × sin(0.6π)

= 8 × 0.951

= 7.608 m

= 8 × 0.951

= 7.608 m

This means the wave point at t = 2 s (or at x = 3 m) is 7.608 m above the equilibrium position.

It is not always possible or suitable to use the general equation of waves as this requires information from many quantities. Also, this process involves complex calculations a user that is not very familiar with math may not understand. Therefore, it is more appropriate to use the simplified equation of waves

v = λ × f

we already known from the previous tutorial in order to calculate any of the missing quantities contained in it. This situation is similar to when we used the simplified equation Fg = m × g for gravitational force instead of the long equation Fg = (G × m1 × m2) / r2.

But first, we must learn how to derive the simplified wave equation from the already confirmed equation of waves. Thus, given that we are discussing for waves whose form does not change in space and time, the argument is constant, i.e.

k × x - ω × t = constant

Taking the time derivative of the above expression, we obtain

k × *dx**/**dt* - ω = 0

k × v - ω = 0

k × v = ω

v =*ω**/**k*

k × v - ω = 0

k × v = ω

v =

Since

ω = *2π**/**T*

= 2π × f

= 2π × f

and

k = *2π**/**λ*

we obtain

v = (*2π**/**T*)/(*2π**/**λ*)

=*λ**/**T*

= λ × f

=

= λ × f

In this way, we confirmed the simplified equation of waves

v = λ × f

obtained in the previous tutorial.

y(x,t) = y_{max} × sin(k × x - ω × t + φ)

is an equation of particles oscillation, while

v = λ × f

is an equation of waves that carry energy, not particles.

In this regard, the equation v = λ × f is officially recognized as the equation of waves.

Just like in SHM, we can find the speed of oscillating particles in a wave by taking the first derivative of y-position with respect to the time. Be careful not confuse the oscillating speed of particles (it lies in the y-direction and is not uniform) with the wave speed (which is horizontal and uniform). Thus,

v_{y}(t) = *dy**/**dt*

=*d[y*_{max} × sin(k × x - ω × t + φ) ]*/**dt*

= y_{max} × ω × cos(k × x - ω × t + φ)

=

= y

Likewise, we obtain for the vertical acceleration of oscillating particles,

a_{y} (t) = *dv**/**dt*

=*-d[y*_{max} × ω × cos(k × x - ω × t + φ) ]*/**dt*

= -y_{max} × ω^{2} × sin(k × x - ω × t + φ)

= -ω^{2} × y(t)

=

= -y

= -ω

A wave oscillates according the equation

y(x,t) = 3 × sin(6π × x - 3π × t)

where x and y are in metres and t is in seconds. Calculate:

- Wavelength
- Period of the wave
- Wave's speed
- Vertical position at t = 2 s of a point of wave that oscillates according the vertical line x = 3 m
- Vertical speed of the above point at t = 2 s

**a)** From the general equation of waves

y(x,t) = y_{max} × sin(k × x - ω × t + φ)

And comparing it with the actual equation

y(x,t) = 3 × sin(6π × x - 3π × t)

we extract the following values:

y_{max} = A = 3 m

k = 6π rad/m

ω = 3π rad/s

k = 6π rad/m

ω = 3π rad/s

and

φ = 0

Thus, giving that

k = *2π**/**λ* = 6π

we obtain for the wavelength λ

λ = *2π**/**6π* = *1**/**3* = 0.33 m

**b)** Since

ω = *2π**/**T* = 3π

we obtain for the period

T = *2π**/**3π* = 0.67 s

**c)** Now we can use the simplified equation of wave

v = λ × f = *λ**/**T*

to find the wave speed. Thus,

v = *0.33 m**/**0.67 s* = 0.5 m/s

**d)** Since the vertical position y of a point of wave at any instant t is given by the equation

y(x,t) = y_{max} × sin(k × x - ω × t + φ)

and since in the specific case this equation is

y(x,t) = 3 × sin(6π × x - 3π × t)

we obtain for x = 3 and t = 2

y(3m,2s) = 3 × sin(6π × 3 - 3π × 2)

= 3 × sin12π

= 0

= 3 × sin12π

= 0

This result means the given point of the wave is at the equilibrium position at t = 2 s.

**e)** Vertical speed is the first derivative of vertical position in respect to the time. Thus, we have

v_{y} = *dy**/**dt*

=*d[3 × sin(6π × x - 3π × t) ]**/**dt*

= 3 × 6π × cos(6π × x - 3π × t)

= 18π × cos(6π × x - 3π × t)

=

= 3 × 6π × cos(6π × x - 3π × t)

= 18π × cos(6π × x - 3π × t)

Thus, for x = 3 and t = 2 we obtain for the vertical speed v_{y}

v_{y} (3m,2s) = 18π × cos(6π × 3 - 3π × 2)

= 18π × cos(12π)

= 18 × 3.14 × 1

= 56.52 m/s

= 18π × cos(12π)

= 18 × 3.14 × 1

= 56.52 m/s

Waves motion is periodical. This means it has many similarities with both circular motion and SHM. The graph of waves motion is sinusoidal, just like in SHM.

Since we are dealing with travelling waves, i.e. with waves that are displaced from the original position, when studying their behavior, we need to consider three quantities instead of two: the x and y coordinates versus time t. Therefore, the equation of waves must be of the type

y(x,t)

Hence, given that the wave equation has a sinusoidal form, we write its general form as

y(x,t) = y_{max} × sin(k × x - ω × t)

where

y_{max} represents the amplitude A of a point of the wave, as it oscillates according the y-axis,

ω represents the angular frequency (ω = 2π / T as usual),

x is the horizontal position,

t is the time of motion, and

y(x, t) is the vertical position at a given instant.

The quantity k is known as the **angular wave number**. It is calculated by

k = *2π**/**λ*

Also, since ω = ** 2π/T**, we can also insert the period T in the wave equation. Therefore, it becomes

y(x,t) = y_{max} × sin(*2π**/**λ* × x - *2π**/**T* × t)

The argument k × x - ω × t is known as the phase.

If there is a phase shift φ from the original position of wave, we insert its value in the equation of waves. Thus, it becomes

y(x,t) = y_{max} × sin(k × x - ω × t + φ)

Since the argument is constant,

k × x - ω × t = constant

we obtain from its first derivation with time

k × *dx**/**dt* - ω=0

k × v - ω = 0

k × v = ω

v =*ω**/**k*

k × v - ω = 0

k × v = ω

v =

and after substituting the known values,

v = λ × f

The equation

y(x,t) = y_{max} × sin(k × x - ω × t + φ)

is an equation of particles oscillation, while

v = λ × f

is an equation of waves that carry energy, not particles.

The speed of oscillating particles in a wave is calculated by taking the first derivative of y-position with respect to the time, i.e.

v_{y} (t) = *dy**/**dt*

=*d[y*_{max} × sin(k × x - ω × t + φ) ]*/**dt*

= y_{max} × ω × cos(k × x - ω × t + φ)

=

= y

Likewise, we obtain for the vertical acceleration of oscillating particles,

a_{y} (t) = *dv**/**dt*

=*-d[y*_{max} × ω × cos(k × x - ω × t + φ) ]*/**dt*

= -y_{max} × ω^{2} × sin(k × x - ω × t + φ)

= -ω^{2} × y(t)

=

= -y

= -ω

*1. A wave particle oscillates according the equation *

y(x,t) = 5 × sin(1.5π × x - 0.5π × t)

where x and y are in metres and t in seconds.

Which of the following is correct from the values of amplitude A, wave number k, and angular frequency ω?

- A = 5 m, k = 1.5 m-1, ω = 0.5 rad/s
- A = 5π m, k = 1.5π m-1, ω = 0.5π rad/s
- A = 5 m, k = 1.5π m-1, ω = 0.5π rad/s
- A = 5 m, k = 1.5π × x m-1, ω = 0.5π × t rad/s

**Correct Answer: C**

*2. What are the speed for a wave whose general equation is *

y(x,t) = 2 × sin(3.5π × x - 2.4π × t)

where x and y are in metres and t is in seconds?

- 0.57 m/s
- 0.69 m/s
- 3.45 m/s
- 0.83 m/s

**Correct Answer: B**

*3. What is the oscillating speed at t = 4 s of the point O for the wave shown in the figure? *

- 0 m/s
- 0.75 m/s
- 2 m/s
- 0.75 π m/s

**Correct Answer: B**

We hope you found this Physics tutorial "General Equation of Waves" useful. If you thought the guide useful, it would be great if you could spare the time to rate this tutorial and/or share on social media, this helps us identify popular tutorials and calculators and expand our free learning resources to support our users around the world have free access to expand their knowledge of physics and other disciplines. In our next tutorial, we expand your insight and knowledge of Waves with our Physics tutorial on Energy and Power of Waves .

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