2 waves are represented as:

y₁= a sin( p + q + r)

y₂ = a cos( p + q)

phase difference between them is?

What if it was

y₁= a cos( p + q + r)

y₂ = a sin( p + q)

These questions have very very simple solutions, but I don't know the method, and know only rudimentary trigonometry.

Please be very clear and explain in the simplest possible way.

Thanks in advance.

To find the phase difference between the two waves you've provided, we need to analyze their mathematical forms. Let's break down each case step by step, using basic trigonometric principles.

Understanding the First Set of Waves

For the first pair of waves, we have:

• y₁ = a sin(p + q + r)
• y₂ = a cos(p + q)

To find the phase difference, we can express both waves in terms of sine functions. Recall that the cosine function can be rewritten in terms of sine:

Cosine to Sine Conversion:

Using the identity: cos(x) = sin(x + π/2), we can rewrite y₂:

• y₂ = a cos(p + q) = a sin(p + q + π/2)

Now, both waves are in sine form:

• y₁ = a sin(p + q + r)
• y₂ = a sin(p + q + r + π/2)

The phase difference (Δφ) between y₁ and y₂ can be calculated as:

Δφ = (p + q + r + π/2) - (p + q + r) = π/2

Exploring the Second Set of Waves

Now, let's look at the second pair of waves:

• y₁ = a cos(p + q + r)
• y₂ = a sin(p + q)

Again, we can convert the cosine function to sine:

• y₁ = a cos(p + q + r) = a sin(p + q + r + π/2)

Now we have:

• y₁ = a sin(p + q + r + π/2)
• y₂ = a sin(p + q)

To find the phase difference here, we calculate:

Δφ = (p + q) - (p + q + r + π/2) = -r - π/2

Summary of Phase Differences

In summary, the phase differences for the two cases are:

• For y₁ = a sin(p + q + r) and y₂ = a cos(p + q): Δφ = π/2
• For y₁ = a cos(p + q + r) and y₂ = a sin(p + q): Δφ = -r - π/2

Understanding these relationships helps in visualizing how waves interact with each other, especially in contexts like sound waves or light waves. If you have any more questions or need further clarification, feel free to ask!