Wave Mechanics Examples: Phase and Phase Difference

This companion entry provides exercises for WM03, Phase and Phase Difference. The exercises are stated first so that they can be attempted without seeing the answers. Complete worked solutions follow in Part II.

Throughout this set, unless a problem explicitly says otherwise, sinusoidal oscillations use the convention

(1)

Under this convention, a larger phase constant represents an oscillator that is further advanced through its cycle at the same clock time.

How to use this problem set

Before calculating, identify whether the problem is asking about an instantaneous phase, a phase difference, or an equivalent phase modulo 2π. For lead/lag problems, write the sign convention first. For time-shift problems, check whether the two oscillators have the same angular frequency before assuming the phase difference is constant.

Part I: Exercises

Exercise 1: Landmarks within one phase cycle

For

state the displacement u at each phase angle:

Then identify which two listed phase angles give the same zero displacement but correspond to different locations within the cycle.

Exercise 2: Converting elapsed time to phase advance

A sinusoidal oscillator has period

Find the phase advance Δ𝜃 during each time interval:

1. 0.20 s,
2. 0.40 s,
3. 0.60 s,
4. 1.00 s.

Express each answer in radians and as a fraction of a complete cycle.

Exercise 3: Equivalent phases modulo 2π

Reduce each phase to an equivalent phase in the interval

1. 5π∕2,
2. −3π∕2,
3. 17π∕6,
4. −11π∕4.

Exercise 4: Equal displacement does not imply equal phase

The following cosine-cycle graph marks two points P and Q with equal displacement.

The points are located at

Answer the following.

1. Show that u_P = u_Q.
2. Find the phase separation 𝜃_Q − 𝜃_P .
3. Reduce that separation to the principal interval (−π,π].
4. Explain why equal displacement alone is insufficient to identify phase.

Exercise 5: Phase difference from two equations

Consider

Determine:

1. ϕ₁ and ϕ₂;
2. Δϕ = ϕ₂ − ϕ₁;
3. which oscillator leads under the cos(ωt + ϕ) convention;
4. whether the unequal amplitudes affect the phase difference.

Exercise 6: Phase difference to time shift

Two same-frequency oscillators have

and constant phase difference

Find:

1. the equivalent time shift Δt;
2. the period T;
3. the fraction of one period represented by the phase difference.

Exercise 7: Time shift to phase difference

Oscillator 2 reaches each corresponding maximum 15 ms earlier than oscillator 1. Both oscillators have frequency

Using the convention u = A cos(ωt + ϕ), determine:

1. the period;
2. the angular frequency;
3. the magnitude of the phase difference;
4. the sign of Δϕ = ϕ₂ − ϕ₁;
5. which oscillator leads.

Exercise 8: Special relative phases

For each phase difference below, state the corresponding fraction of one cycle and give the usual qualitative description.

1. 0,
2. π∕2,
3. π,
4. 3π∕2,
5. 2π.

For part (d), also give the equivalent principal phase difference in (−π,π].

Exercise 9: Principal phase difference

Reduce each phase difference to the interval

1. 7π∕4,
2. −5π∕3,
3. 13π∕6,
4. −9π∕4.

For each result, state whether oscillator 2 leads or lags oscillator 1 under the cos(ωt + ϕ) convention.

Exercise 10: Read lead and lag from a graph

The graph below compares two equal-frequency sinusoids.

The solid curve is

and the dashed curve is shifted by one sixth of a period toward earlier time.

1. Which waveform leads?
2. What is the phase difference Δϕ = ϕ₂ − ϕ₁?
3. Write an equation for u₂(t).
4. If T = 0.30 s, what is the time lead?

Exercise 11: Phase difference with unequal amplitudes

Consider

Find the phase difference and state which oscillator leads. Then explain why the amplitude ratio 9∕2 does not enter the phase-difference calculation.

Exercise 12: Phase drift for unequal frequencies

Two oscillators have phases

1. Find Δ𝜃(t) = 𝜃₂(t) − 𝜃₁(t).
2. Find the initial phase difference Δ𝜃(0).
3. Find the relative phase at t = 1 s.
4. Does oscillator 2 gain phase or lose phase relative to oscillator 1?
5. Why is it incorrect to describe the two oscillators as having one permanent phase lead?

The next graph illustrates the general idea of phase drift.

Exercise 13: Sign conventions matter

Two authors describe an otherwise identical harmonic motion using different forms:

Suppose ϕ > 0 and ψ > 0.

1. Under Author A’s convention, does positive ϕ shift a maximum to earlier or later time relative to A cos(ωt)?
2. Under Author B’s convention, does positive ψ shift a maximum to earlier or later time?
3. Explain why the statement “positive phase always means lead” is unsafe unless the sign convention is given.

Exercise 14: Challenge—infer phase from measured maxima

Two sensors record same-frequency sinusoidal signals at

A maximum of sensor 1 occurs at

while the corresponding maximum of sensor 2 occurs at

Assume the relative phase difference is constant and use

Determine:

1. which signal leads;
2. the time lead;
3. the period and angular frequency;
4. the phase difference in radians;
5. the phase difference in degrees;
6. an equation for sensor 2 if sensor 1 is represented by
   

   and sensor 2 has amplitude A₂.

Part II: Complete Worked Solutions

Solution 1: Landmarks within one phase cycle

For u = A cos 𝜃,

𝜃 = 0	: u = A,	(2)
𝜃 =	: u = 0,	(3)
𝜃 = π	: u = −A,	(4)
𝜃 =	: u = 0,	(5)
𝜃 = 2π	: u = A.	(6)

Thus π∕2 and 3π∕2 both give zero displacement, but they are separated by half a cycle and are not the same phase location.

Key idea. Displacement alone does not specify phase. Most displacement values occur twice during one cosine cycle.

Solution 2: Converting elapsed time to phase advance

Use

1. For Δt = 0.20 s,
   

   so

   

   This is one quarter cycle.

2. For 0.40 s,
   

   This is one half cycle.

3. For 0.60 s,
   

   This is three quarters of a cycle.

4. For 1.00 s,
   

   so

   

   This is one and one quarter cycles.

Notice that accumulated phase can exceed 2π. It may later be reduced modulo 2π if only the location within the current cycle is needed.

Solution 3: Equivalent phases modulo 2π

We add or subtract integer multiples of 2π until the result lies in 0 ≤ 𝜃 < 2π.

1. 
2. 
3. 
4. Add 4π:
   

Solution 4: Equal displacement does not imply equal phase

The marked phases are

1. 

   while

   

   Therefore

   
2. 
3. The value 4π∕3 lies outside (−π,π]. Subtract 2π:
   

   Thus the principal signed separation is

   
4. The cosine function is not one-to-one over a full cycle. The same vertical displacement can occur at different phase angles, and those points can correspond to opposite directions of motion. Therefore displacement alone does not determine phase.

Solution 5: Phase difference from two equations

The equations are

1. 

2. Δϕ	= ϕ2 − ϕ1	(7)
   	= −	(8)
   	= −	(9)
   	= .	(10)

3. Since Δϕ > 0 under the +ϕ cosine convention,
   
4. No. Amplitude controls vertical scale. Phase difference depends on the phase angles, not on A₁ or A₂.

Solution 6: Phase difference to time shift

Given

1. Δt	=	(11)
   	=	(12)
   	= s	(13)
   	= 0.030 s .	(14)

2. 
3. 

   Equivalently,

   

   Thus the phase difference represents three tenths of a cycle.

Solution 7: Time shift to phase difference

The frequency is 10 Hz and oscillator 2 reaches each corresponding maximum 15 ms earlier.

1. 
2. 
3. Convert the time lead:
   

   Then

   
4. Oscillator 2 occurs earlier, so under the +ϕ convention it is advanced. Therefore
   
5. 

Solution 8: Special relative phases

1. 

   The oscillators are

   in phase .

2. 

   This is a

   quarter-cycle separation .

3. 

   This is a half-cycle separation, commonly called

   
4. 

   This is three quarters of a cycle. In the principal interval,

   

   so it can equivalently be described as a lag of one quarter cycle.

5. 

   One full cycle returns to the same relative phase, so the oscillators are again

   equivalent in phase .

Solution 9: Principal phase difference

Reduce each value to −π < Δϕ ≤ π.

1. 

   Thus

   

   so oscillator 2 lags by π∕4.

2. 

   Thus

   

   so oscillator 2 leads by π∕3.

3. 

   Oscillator 2 leads by π∕6.

4. 

   Thus oscillator 2 lags by π∕4.

Common error. A large positive angle and a small negative angle can describe the same relative phase. Reducing to a stated principal interval removes that ambiguity.

Solution 10: Read lead and lag from a graph

The dashed waveform reaches the corresponding maximum one sixth of a period earlier than the solid waveform.

1. The dashed waveform, u₂, leads.
2. One sixth of a full phase cycle is
   

   Because u₂ leads under the +ϕ convention, the sign is positive.

3. 
4. 

Solution 11: Phase difference with unequal amplitudes

The phase constants are

Thus

Δϕ	= ϕ2 − ϕ1	(15)
	= −	(16)
	= .	(17)

Therefore oscillator 2 leads oscillator 1 by one quarter cycle.

The amplitudes 2 and 9 only scale the vertical displacement. They do not change the arguments of the cosine functions, so they do not enter the phase-difference calculation.

Solution 12: Phase drift for unequal frequencies

Given

subtract:

Δ𝜃(t)	= 𝜃2 − 𝜃1	(18)
	= (7π − 6π)t +	(19)
	= πt −.	(20)

Therefore

1. The expression above is the required relative phase.
2. At t = 0,
   
3. At t = 1 s,
   
4. Because
   

   oscillator 2 continually gains phase relative to oscillator 1.

5. The phase difference contains the time-dependent term πt. Therefore it is not constant, so no single permanent phase lead describes the relationship for all time.

Solution 13: Sign conventions matter

1. For
   

   a maximum satisfies

   

   near the reference maximum, so

   

   For ϕ > 0, the maximum occurs at an earlier time. Thus positive ϕ represents a lead under this convention.

2. For
   

   the corresponding maximum satisfies

   

   so

   

   For ψ > 0, the maximum occurs later in time. Under this convention a positive parameter produces a lag relative to A cos(ωt).

3. The sign attached to the phase parameter inside the cosine argument changes the verbal lead/lag interpretation. Therefore the equation must be inspected before assigning meaning to a positive phase parameter.

Solution 14: Challenge—infer phase from measured maxima

The frequency is

The corresponding maxima occur at

1. Sensor 2 reaches the corresponding maximum first, so
   
2. The time lead magnitude is
   
3. 

   while

   
4. Under the +ϕ convention, the earlier signal has positive relative phase:

   Δϕ	= ωΔt	(21)
   	= (10π)(0.030)	(22)
   	= 0.30π = .	(23)

5. Convert radians to degrees:
   
6. If sensor 1 is
   

   then sensor 2 has phase constant

   

   Thus

   

Reasonableness check. A 0.030 s lead out of a 0.20 s period is

of a cycle. Multiplying 0.15 by 360^∘ gives 54^∘, consistent with the phase calculation.

Summary of skills practiced

After completing this set, you should be able to:

• identify standard phase locations within a cosine cycle;
• convert elapsed time into phase advance;
• reduce phase angles modulo 2π;
• explain why equal displacement does not imply equal phase;
• calculate Δϕ = ϕ₂ − ϕ₁ for equal-frequency sinusoids;
• interpret phase lead and lag under a stated sign convention;
• convert between phase difference and time shift;
• choose a principal relative phase;
• recognize that amplitude and phase are independent descriptors;
• calculate phase drift when angular frequencies differ;
• diagnose lead/lag changes caused by different phase-sign conventions.

The next main lesson, WM04, moves from temporal oscillation to spatial variation and introduces wavelength as the spatial analogue of period.