This companion article provides self-study exercises for WM01, Oscillation at One Point. The problems are stated first so that they can be attempted without seeing the answers. Complete worked solutions follow in Part II. The exercises remain intentionally within the WM01 foundation: equilibrium, signed displacement, distance traveled, amplitude, cycles, period, frequency, and interpretation of a one-point time history u(t). Angular frequency, phase, sinusoidal formulas, wavelength, and traveling waves are reserved for later lessons.

How to use this problem set

Try each exercise before reading Part II. Write the physical quantity being sought, include units at every numerical step, and check whether the result is physically reasonable. For graph problems, identify the equilibrium line first and remember that a complete cycle requires repetition of the motion, not merely repetition of one displacement value.

Part I: Exercises

Exercise 1: Equilibrium and signed displacement

A small cart oscillates along a straight track. Its equilibrium position is assigned the coordinate u = 0. The positive direction is to the right.

1. What does u = +4.0 cm mean physically?
2. What does u = −2.5 cm mean physically?
3. What is the displacement when the cart is exactly at equilibrium?
4. If the positive axis were reversed, what would happen to the signs of the two nonzero coordinates above? Would the physical locations change?

Exercise 2: Displacement versus distance traveled

A mass starts at equilibrium, moves successively through

Find:

1. the final displacement;
2. the total distance traveled.

Explain in one sentence why the two answers are different.

Exercise 3: Amplitude and peak-to-peak range

An oscillator moves between the extreme displacements

Find:

1. the amplitude A;
2. the peak-to-peak range;
3. the displacement when the oscillator passes through equilibrium.

Exercise 4: Reading amplitude, period, and frequency from a graph

The following time-history graph will be used to read the amplitude, period, and frequency of a periodic oscillator.

Determine:

1. the equilibrium value;
2. the amplitude A;
3. the period T;
4. the frequency f;
5. how many complete cycles occur between t = 0.1 s and t = 0.9 s.

Exercise 5: Period to frequency

A floating sensor completes one vertical oscillation every 0.80 s.

1. What is its period?
2. What is its frequency in hertz?
3. In words, what does the numerical value of the frequency mean?

Exercise 6: Frequency to period

A mechanical vibrator operates at 25 Hz.

1. Find the period in seconds.
2. Express the period in milliseconds.

Exercise 7: Counting cycles in a time interval

An oscillator has frequency

How many complete cycles occur in:

1. 1 s;
2. 3 s;
3. 0.50 s?

Exercise 8: Time required for a specified number of cycles

A system oscillates at 4 Hz.

1. What is its period?
2. How long does one cycle take?
3. How long do 10 complete cycles take?
4. How long do 100 complete cycles take?

Exercise 9: Comparing two oscillators

Oscillator A has period

while oscillator B has frequency

Determine:

1. the frequency of A;
2. the period of B;
3. which oscillator completes more cycles in 10 s;
4. how many cycles each completes in that interval.

Exercise 10: Fundamental period versus another repeat time

A periodic motion has a smallest positive repeat time of

For each time interval below, state whether it is also a repeat time of the motion and whether it is the fundamental period:

Then find the frequency.

Exercise 11: Same displacement, different motion

The following graph marks two instants P and Q at which the oscillator has the same displacement.

Answer the following without introducing phase notation.

1. Is u_P = u_Q?
2. Is the direction of motion the same at P and Q?
3. Has the complete mechanical state necessarily repeated merely because the displacement is equal?
4. Why can measuring a period by choosing two arbitrary equal-displacement crossings give the wrong answer?

Exercise 12: Unit conversion and reciprocal reasoning

Complete the following conversions.

1. If T = 2.0 ms, find f in hertz.
2. Express the answer to part (a) in kilohertz.
3. If f = 1.25 kHz, find T in seconds.
4. Express the answer to part (c) in milliseconds.

Exercise 13: Mixed specification problem

A laboratory shaker is specified to have amplitude

and frequency

Find:

1. the maximum positive displacement;
2. the maximum negative displacement;
3. the peak-to-peak range;
4. the period;
5. the number of complete cycles in 12 s.

Exercise 14: Challenge—reason from observations

During a test, an engineer notes that identical maxima of an oscillator occur at

Assume these are consecutive maxima of a periodic motion.

1. Determine the period from the observations.
2. Determine the frequency.
3. How many cycles occur between the first and fourth listed maxima?
4. Why is the elapsed time between the first and fourth maxima not four periods?

Part II: Complete Worked Solutions

Solution 1: Equilibrium and signed displacement

The coordinate is measured from equilibrium and the positive direction is rightward.

1. 

   means the cart is 4.0 cm to the right of equilibrium.

2. 

   means the cart is 2.5 cm to the left of equilibrium.

3. At equilibrium,
   
4. Reversing the positive axis reverses the signs of all signed coordinates. The same physical points would therefore be described by −4.0 cm and +2.5 cm, respectively. The coordinate description changes, but the physical locations do not.

Common error. A negative displacement does not mean a negative distance or an impossible location. It only indicates the side of the chosen origin.

Solution 2: Displacement versus distance traveled

The motion is

1. The final displacement is simply the final coordinate relative to equilibrium:
   
2. Distance traveled must be accumulated along every segment.

   From 0 to +3 cm:

   

   From +3 cm to −2 cm:

   

   From −2 cm to +1 cm:

   

   Therefore

   

Displacement depends only on the final position relative to equilibrium; total distance records the entire path traveled.

Solution 3: Amplitude and peak-to-peak range

The extremes are symmetric about equilibrium:

1. The amplitude is the maximum magnitude of displacement:
   
2. The peak-to-peak range is the distance between the two extremes:
   

   Equivalently, the range is 2A.

3. At equilibrium,
   

Common error. The amplitude is 7.5 mm, not 15.0 mm. The latter is the peak-to-peak range.

Solution 4: Reading amplitude, period, and frequency from a graph

Refer to the time-history graph included earlier in this entry.

1. The motion is centered on
   

   so this is the equilibrium value.

2. The extrema are +3 mm and −3 mm, giving
   
3. Consecutive maxima occur at approximately
   

   The separation between consecutive identical maxima is therefore

   
4. Using f = 1∕T,
   
5. The interval from 0.1 s to 0.9 s is
   

   With T = 0.4 s,

   

Common error. Measuring from a maximum to the following minimum would give 0.20 s, only half of the full repeat time shown by the graph.

Solution 5: Period to frequency

1. The statement “one oscillation every 0.80 s” directly gives
   
2. 

   Hence

   
3. A frequency of 1.25 Hz means the oscillator completes 1.25 cycles per second on average for this periodic motion.

Solution 6: Frequency to period

Given

we use

1. 

   Thus

   
2. Since 1 s = 1000 ms,
   

Solution 7: Counting cycles in a time interval

A frequency of 6 Hz means 6 cycles occur each second. Thus

1. 
2. 
3. 

The units cancel as they should: s⁻¹s = 1, leaving a dimensionless cycle count.

Solution 8: Time required for a specified number of cycles

The frequency is

1. 
2. One cycle takes one period, so
   
3. Ten cycles require
   
4. One hundred cycles require
   

Solution 9: Comparing two oscillators

For oscillator A,

For oscillator B,

1. 
2. 
3. Since 5.0 Hz > 2.0 Hz, oscillator A completes more cycles in any equal time interval.
4. In 10 s,
   

   while

   

Solution 10: Fundamental period versus another repeat time

The smallest positive repeat time is given as

Therefore every positive integer multiple nT is also a repeat time.

The frequency is

To an appropriate number of significant figures,

Key idea. A periodic function has many repeat times, but the period normally means the smallest positive repeat time.

Solution 11: Same displacement, different motion

At both marked points, the graph lies on the same horizontal level.

1. Yes:
   
2. No. At P the graph is decreasing, while at Q it is increasing. For a mechanical oscillator this means the velocity has opposite sign at the two instants.
3. No. Equal displacement alone is not enough to establish that the complete mechanical state has repeated. The direction or rate of motion must also agree.
4. A periodic oscillator can pass through the same displacement more than once during one cycle. Choosing two equal-displacement crossings without checking direction can therefore select a time separation smaller than one full period.

This is one reason the safest graphical period measurement is often maximum-to-next-maximum or minimum-to-next-minimum.

Solution 12: Unit conversion and reciprocal reasoning

1. Convert milliseconds to seconds first:
   

   Then

   
2. Since 1000 Hz = 1 kHz,
   
3. 

   Hence

   

   Thus

   
4. Converting to milliseconds,
   

Reasonableness check. A frequency in kilohertz should correspond to a period on the order of milliseconds or less, which our answer does.

Solution 13: Mixed specification problem

Given

1. The maximum positive displacement is
   
2. The maximum negative displacement is
   
3. The peak-to-peak range is
   
4. 
5. 

Notice that amplitude and frequency describe different aspects of the motion: A specifies its size, while f specifies how rapidly the cycle repeats.

Solution 14: Challenge—reason from observations

The listed maxima occur at

Because they are consecutive identical maxima, the time between any neighboring pair is one period.

1. 

   Checking the next pair,

   

   confirms the same repeat time. Therefore

   
2. 
3. From the first listed maximum to the fourth, the oscillator goes through the intervals
   

   These are three complete periods, so

   
4. Four listed maxima contain only three intervals between them. Numerically,
   

   Therefore the elapsed time is three periods, not four.

Common counting error. When counting events, distinguish the number of marked points from the number of intervals between those points.

What this set prepares you for

WM01E1 has used only the timing and displacement ideas of WM01. The next main lesson, WM02, gives a specific mathematical shape to smooth periodic motion and introduces angular frequency and phase. Those new ideas will allow the same graphical reasoning used here to be expressed algebraically.

Summary of skills practiced

After completing this set, you should be able to:

• interpret equilibrium and signed displacement;
• distinguish displacement from distance traveled;
• distinguish amplitude from peak-to-peak range;
• identify amplitude and period from a time-history graph;
• convert between period and frequency;
• count cycles over a specified time;
• distinguish a fundamental period from its multiples;
• explain why equal displacement does not necessarily mean the complete state has repeated;
• use units and reciprocal reasoning to check oscillation calculations.