The diagram below represents a periodic wave. Which point on the wave is

90° out of phase with point P?

A

B

C

D

Point P lies on the equilibrium (central) line on the falling part of the first crest.

• A is where that crest starts from the equilibrium line.

• B is the trough (minimum).

• C is on the next crest, above the equilibrium line.

• D is on the equilibrium line at the end of that crest.

Key Concept: Phase Difference on a Wave

For a sinusoidal wave:

• One full wavelength corresponds to a phase change of 360∘360∘.

• A quarter of a wavelength corresponds to 90∘90∘.

So, two points are 90 degrees out of phase if they are separated by a horizontal distance of one‑quarter of a wavelength (λ4)(4λ) along the direction of the wave.

On a standard sine wave:

• From equilibrium (rising) to crest = λ44λ =90∘=90∘.

• From crest to equilibrium (falling) = another λ44λ.

• From equilibrium (falling) to trough = another λ44λ.

This pattern continues periodically.

Step‑by‑Step Reasoning for Point P

From the diagram:

1. The wave starts from A on the equilibrium line, rises to a crest, and then falls down.

2. P is on the downward‑sloping part of that crest, crossing the equilibrium line.

3. Moving from P further along the wave, the next characteristic point is the trough at B.

The horizontal distance from P to B visually corresponds to one‑quarter of a wavelength, because:

• From equilibrium (falling) at P to the trough at B represents going from a phase like 0∘0∘ (crossing downwards) to −90∘−90∘.

• That is exactly a phase change of 90∘90∘.

Therefore, point B is 90 degrees out of phase with point P.

Correct answer: B) B

Explanation of Each Option

Option A: Point A

• A is on the equilibrium line at the beginning of the crest, where the wave starts to rise.

• The path from A to P is less than a quarter of a wavelength (they are on the same side of the same crest).

• Their phase difference is less than 90∘90∘ and they are not separated by the characteristic quarter‑cycle distance.

So, point A is not 90 degrees out of phase with P.

Option B: Point B (Correct)

• B is the lowest point of the trough immediately after P.

• Going from P (equilibrium, moving downward) to B (minimum displacement) is exactly a quarter of a cycle of motion.

• A quarter of a cycle corresponds to a phase difference of 90∘90∘.

Hence, point B is 90 degrees out of phase with point P.

Option C: Point C

• C lies on the next crest, above the equilibrium line.

• The distance from P to C is more than λ44λ and in fact corresponds to a larger phase change (closer to 270∘270∘ or $3\pi /2$ depending on the reference).

• They are still part of the same periodic wave, but not separated by exactly a quarter wavelength.

So, point C is not 90 degrees out of phase with P.

Option D: Point D

• D is located on the equilibrium line at the end of the next crest.

• The phase difference between P and D is approximately a full wavelength =360∘=360∘.

• Points separated by a full wavelength are in phase, not 90∘90∘ out of phase.

So, point D is in phase with P, not 90 degrees out of phase.

Final Answer

The point on the wave that is 90 degrees out of phase with point P is:

B) B