Wave Mechanics Examples: Sinusoidal Oscillation

This companion entry provides self-study exercises for WM02, Sinusoidal Oscillation. All exercises are stated before the solutions so that the problems can be attempted independently. The set develops the relationships among period, frequency, angular frequency, phase angle, and the sinusoidal form

(1)

The problems remain at one spatial point. Wavelength, wavenumber, spatial phase, and traveling waves are intentionally excluded.

Useful relationships

The following WM02 relationships are sufficient for every exercise in this entry:

f	= ,	(2)
ω	= 2πf = ,	(3)
𝜃(t)	= ωt + ϕ,	(4)
u(t)	= A cos(ωt + ϕ),	(5)
u(0)	= A cos ϕ.	(6)

One complete cycle corresponds to an angular advance of 2π radians.

Part I: Exercises

Exercise 1: Cycles and radians

Convert each quantity.

1. How many radians correspond to one complete cycle?
2. How many radians correspond to one quarter of a cycle?
3. How many radians correspond to 1.5 cycles?
4. How many cycles correspond to a phase advance of 5π radians?

Exercise 2: Period, frequency, and angular frequency

An oscillator has period

(7)

Determine:

1. the frequency f;
2. the angular frequency ω;
3. the number of radians of phase accumulated in 0.50 s.

Exercise 3: Start from angular frequency

A sinusoidal oscillator has

(8)

Find:

1. its frequency in hertz;
2. its period in seconds;
3. the phase advance during one period.

Exercise 4: Read a sinusoid from a graph

The graph below shows a sinusoidal displacement at one point.

Determine:

1. the amplitude A;
2. the period T;
3. the frequency f;
4. the angular frequency ω;
5. a cosine equation for the graph using the simplest phase constant.

Exercise 5: Construct a sinusoidal equation from frequency

An oscillator has amplitude

(9)

and frequency

(10)

At t = 0 it is at maximum positive displacement.

1. Find ω.
2. Find T.
3. Write u(t) in cosine form.

Exercise 6: Construct a sinusoidal equation from period and phase

A displacement has amplitude

(11)

period

(12)

and phase constant

(13)

Determine f, ω, and the cosine equation u(t).

Exercise 7: Evaluate a sinusoid at special times

Consider

(14)

where t is measured in seconds.

Find u(t) at

(15)

Then state the period.

Exercise 8: Phase angle and displacement

Consider

(16)

For each time below, determine both the phase angle 𝜃(t) and the displacement u(t):

1. t = 0;
2. t = 0.10 s;
3. t = 0.20 s.

Exercise 9: Initial phase and initial displacement

For the general form

(17)

find u(0) for each phase constant:

1. ϕ = 0;
2. ϕ = π∕2;
3. ϕ = π;
4. ϕ = 3π∕2.

Which cases begin at an extreme displacement, and which begin at equilibrium?

Exercise 10: Cosine and sine descriptions

Use the identities

(18)

to rewrite:

1. 3 cos(2πt) in sine form;
2. 2 sin(4πt) in cosine form.

Explain why the rewritten equations represent the same physical time histories.

Exercise 11: Compare two sinusoidal equations

Consider

u1(t)	= 2.0 cm cos(6πt),	(19)
u2(t)	= 5.0 cm cos .	(20)

For each oscillator identify A, ω, f, T, and ϕ. Then answer:

1. Do the oscillators have the same amplitude?
2. Do they have the same repetition rate?
3. Do they begin at the same point in their cycles?

Exercise 12: Determine the phase constant from a starting point

The graph below has amplitude 4.0 cm and period 1.0 s. At t = 0 the displacement is +2.0 cm and the displacement immediately begins to decrease.

Using

(21)

determine a phase constant in the interval 0 ≤ ϕ < 2π and write the complete equation.

Exercise 13: Diagnose an incorrect equation

A student is told that an oscillator has amplitude 2.0 cm, frequency 3.0 Hz, and phase constant π. The student writes

(22)

Is this equation correct? If not, identify the error and write the correct equation.

Exercise 14: Challenge—build the equation from observations

An oscillator completes eight cycles in 2.0 s. Its amplitude is 7.0 mm. At t = 0 it is at equilibrium and immediately afterward its displacement becomes negative.

Determine:

1. the frequency;
2. the period;
3. the angular frequency;
4. an appropriate phase constant in 0 ≤ ϕ < 2π;
5. the complete cosine equation u(t).

Part II: Complete Worked Solutions

Solution 1: Cycles and radians

One complete cycle corresponds to 2π radians.

1. (23)

2. One quarter of a cycle is

   (24)

3. For 1.5 cycles,

   (25)

4. The number of cycles is

   (26)

Common error. Do not confuse radians with cycles. The conversion factor is 2π radians per cycle.

Solution 2: Period, frequency, and angular frequency

Given

(27)

we first compute the frequency:

(28)

Then

(29)

During 0.50 s, the phase advance is

Δ𝜃	= ωΔt	(30)
	= (8π rad/s)(0.50 s)	(31)
	= 4π rad .	(32)

This is two complete cycles, which is consistent with a period of 0.25 s.

Solution 3: Start from angular frequency

Given

(33)

use f = ω∕(2π):

(34)

The period is

(35)

By definition, the phase advance over one period is

(36)

Solution 4: Read a sinusoid from a graph

The graph reaches +3 cm and −3 cm, so

(37)

Consecutive positive maxima occur at t = 0 and t = 0.50 s, giving

(38)

Therefore

(39)

and

(40)

Because the graph begins at maximum positive displacement, the simplest phase constant is ϕ = 0. Thus

(41)

Common error. The time from a maximum to the next minimum is half a period, not a full period.

Solution 5: Construct a sinusoidal equation from frequency

The angular frequency is

(42)

The period is

(43)

Maximum positive displacement at t = 0 corresponds to the simplest choice ϕ = 0. Therefore

(44)

Solution 6: Construct a sinusoidal equation from period and phase

The frequency is

(45)

The angular frequency is

(46)

Substituting the given amplitude and phase constant,

(47)

At t = 0, this gives u(0) = 0, as expected from cos(−π∕2) = 0.

Solution 7: Evaluate a sinusoid at special times

The equation is

(48)

At the requested times:

u(0)	= 4.0 cm cos 0 = +4.0 cm ,	(49)
u(0.25)	= 4.0 cm cos = 0 ,	(50)
u(0.50)	= 4.0 cm cos(π) = −4.0 cm ,	(51)
u(0.75)	= 4.0 cm cos = 0 ,	(52)
u(1.00)	= 4.0 cm cos(2π) = +4.0 cm .	(53)

Since ω = 2π rad/s,

(54)

Solution 8: Phase angle and displacement

Here

(55)

At t = 0,

(56)

At t = 0.10 s,

𝜃(0.10)	= 5π(0.10) +	(57)
	= π,	(58)

so

(59)

At t = 0.20 s,

𝜃(0.20)	= 5π(0.20) +	(60)
	= ,	(61)

so

(62)

Solution 9: Initial phase and initial displacement

At t = 0,

(63)

Therefore

ϕ = 0	: u(0) = +A ,	(64)
ϕ =	: u(0) = 0 ,	(65)
ϕ = π	: u(0) = −A ,	(66)
ϕ =	: u(0) = 0 .	(67)

The cases ϕ = 0 and ϕ = π begin at extreme displacements. The cases ϕ = π∕2 and ϕ = 3π∕2 begin at equilibrium.

Important point. Equal displacement does not by itself identify the complete state. The two equilibrium cases occupy different locations within the repeating cycle.

Solution 10: Cosine and sine descriptions

Using

(68)

we obtain

(69)

Using

(70)

we obtain

(71)

These are not different motions. They are different mathematical descriptions of the same time histories because sine and cosine differ only by a phase shift.

Solution 11: Compare two sinusoidal equations

For

(72)

we identify

(73)

Thus

(74)

For

(75)

we identify

(76)

Hence

(77)

The amplitudes are different, the repetition rates are the same, and the initial phases are different. Therefore the oscillators do not begin at the same point in their cycles.

Solution 12: Determine the phase constant from a starting point

The period is 1.0 s, so

(78)

and

(79)

At t = 0,

(80)

Using u(0) = 2.0 cm and A = 4.0 cm,

(81)

so

(82)

Within 0 ≤ ϕ < 2π, the two angles with cosine 1∕2 are

(83)

The graph decreases immediately after t = 0. As phase advances from π∕3, cosine decreases toward zero and then becomes negative. As phase advances from 5π∕3, cosine increases toward 1. Therefore the graph selects

(84)

The complete equation is

(85)

Solution 13: Diagnose an incorrect equation

The student used the numerical frequency f = 3.0 Hz directly as the coefficient of t. The cosine argument requires angular frequency, not cycles per second.

The correct angular frequency is

(86)

Therefore the correct equation is

(87)

Common error. The forms cos(ft) and cos(ωt) are not interchangeable when f is measured in hertz. They differ by a factor of 2π.

Solution 14: Challenge—build the equation from observations

Eight cycles occur in 2.0 s, so

(88)

The period is

(89)

The angular frequency is

(90)

At t = 0 the oscillator is at equilibrium, so

(91)

Within one cycle, the two simplest possibilities are

(92)

Immediately after π∕2, cosine becomes negative. Immediately after 3π∕2, cosine becomes positive. The observation that the displacement becomes negative therefore selects

(93)

With A = 7.0 mm, the complete equation is

(94)

This result also provides a useful physical check: at t = 0 the cosine is zero, and a small positive increase in phase from π∕2 makes the cosine negative.

Summary of skills practiced

These exercises reinforce the sequence

(95)

The main conceptual distinction is that f measures cycles per second while ω measures radians of phase advance per second. The amplitude controls the vertical scale, and the phase constant selects the starting point within the repeating sinusoidal cycle.