# How can the direction of propagation help in determining the phase of a wave?

• Philip551
In summary, the conversation discusses finding the wavefunction of a sinusoidal wave propagating on a string, given certain parameters. The equations for k and omega are used to find the phase, but there is not enough information to determine the exact value of phi without another initial condition. The direction of propagation is also important to consider in the calculation.
Philip551
Homework Statement
Find the wavefunction of a sinusoidal wave that propagates on a string towards the negative direction of the x-axis given that ##y_{max}## = 8cm, f = 3Hz, ##\lambda## = 80cm and that y(0,t)=0 at t=0.
Relevant Equations
$$y(x,t)= y_{max} sin(kx- \omega t + \phi)$$
Using the equation above I know that I have to find parameters k ##\omega## and ##\phi##.

$$k = \frac{2\pi}{\lambda}$$

and

$$\omega = 2\pi f$$

The problem I've been having is how you would go about finding ##\phi## since by solving:

$$y(0,0)=0 \rightarrow sin(\phi)=0 \rightarrow \phi = 0, \pi$$

you get two different possible values for phi. How would you decide which one is correct without another initial condition?

Why is ##y(0,0)=0##? Note that you should write ##y(x,t)=y_{max}\sin(kx-\omega t+\phi)## not ##y(x,t)=y_{m}ax\sin(kx-\omega t+\phi)##. The amplitude in LaTeX should be (without the delimiters) y_{max} not y_max.

kuruman said:
Why is ##y(0,0)=0##? Note that you should write ##y(x,t)=y_{max}\sin(kx-\omega t+\phi)## not ##y(x,t)=y_{m}ax\sin(kx-\omega t+\phi)##. The amplitude in LaTeX should be (without the delimiters) y_{max} not y_max.
I understood that y(0,t)=0 at t =0 (from the homework statement)is the same as y(0,0)=0

Philip551 said:
Homework Statement:: Find the wavefunction of a sinusoidal wave that propagates on a string towards the negative direction of the x-axis given that ##y_{max}## = 8cm, f = 3Hz, ##\lambda## = 80cm and that y(0,t)=0 at t=0.
Relevant Equations:: $$y(x,t)= y_{max} sin(kx- \omega t + \phi)$$

Using the equation above I know that I have to find parameters k ##\omega## and ##\phi##.

$$k = \frac{2\pi}{\lambda}$$

and

$$\omega = 2\pi f$$

The problem I've been having is how you would go about finding ##\phi## since by solving:

$$y(0,0)=0 \rightarrow sin(\phi)=0 \rightarrow \phi = 0, \pi$$

you get two different possible values for phi. How would you decide which one is correct without another initial condition?
Yes, there is not enough information to determine the phase exactly.
But you seem to have ignored the info about the direction of propagation.

haruspex said:
Yes, there is not enough information to determine the phase exactly.
But you seem to have ignored the info about the direction of propagation.
That is what I though.

I forgot to include that. It should be:

$$y(x,t) = y_{max} sin(kx+\omega t + \phi)$$

haruspex

## 1. What is a phase of a wave?

The phase of a wave refers to the position of a point on the wave at a specific time. It is used to describe the relationship between two waves or different points on the same wave.

## 2. How is the phase of a wave measured?

The phase of a wave is measured in degrees or radians. It is usually measured relative to a reference point, such as the starting point of the wave or another wave.

## 3. What is the difference between phase and frequency?

While phase refers to the position of a point on a wave, frequency refers to the number of complete cycles of a wave that occur in a given time. Phase and frequency are related, but they are different properties of a wave.

## 4. How can the phase of a wave be changed?

The phase of a wave can be changed by altering the starting point or reference point of the wave, or by introducing a phase shift using a device such as a phase shifter.

## 5. Why is finding the phase of a wave important?

Finding the phase of a wave is important in many fields, including physics, engineering, and telecommunications. It allows for the analysis and manipulation of waves, which is crucial in understanding and utilizing various phenomena and technologies.

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