Playing melodies with tone() in Arduino - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to play a tone on pin 8 at 440 Hz.

Arduino

tone(8, [1]);

Drag options to blanks, or click blank then click option'

A440

B8

C1000

Dtone

Attempts:

3 left

💡 Hint

Common Mistakes

Using the pin number as the frequency.

Using the function name as a parameter.

The tone() function plays a sound at the specified frequency in hertz. 440 Hz is the standard A note.

2fill in blank

medium

Complete the code to stop the tone playing on pin 8.

Arduino

noTone([1]);

Drag options to blanks, or click blank then click option'

A440

B1000

Ctone

D8

Attempts:

3 left

💡 Hint

Common Mistakes

Passing frequency instead of pin number.

Passing the function name.

The noTone() function stops the tone on the specified pin. You must give the pin number.

3fill in blank

hard

Fix the error in the code to play a 500 Hz tone for 1000 milliseconds on pin 9.

Arduino

tone(9, [1], 1000);

Drag options to blanks, or click blank then click option'

Atone

B9

C500

D1000

Attempts:

3 left

💡 Hint

Common Mistakes

Using the pin number as frequency.

Using the duration as frequency.

The second argument is the frequency in hertz. 500 is the correct frequency to play.

4fill in blank

hard

Fill both blanks to create a dictionary of notes with their frequencies and play the note 'C' on pin 7.

Arduino

int notes[] = [1];
tone(7, notes[[2]]);

Drag options to blanks, or click blank then click option'

A262, 294, 330

B0

C1

D2

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong indices for notes.

Not using correct frequencies.

The array contains frequencies for notes C, D, E. Index 0 is for C, so playing notes[0] plays C.

5fill in blank

hard

Fill all three blanks to create a loop that plays notes from an array on pin 6 with 500 ms delay between notes.

Arduino

int melody[] = [1];
for (int i = 0; i < [2]; i++) {
  tone(6, melody[[3]]);
  delay(500);
}

Drag options to blanks, or click blank then click option'

A262, 294, 330, 349

B4

Ci

D3

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong loop count.

Using fixed index instead of loop variable.

The array has 4 notes. The loop runs 4 times. Using i as index plays each note in order.

Practice

1. What does the tone() function do in Arduino?

easy

A. It reads the value from a sensor.

B. It controls the brightness of an LED.

C. It plays a sound at a specific frequency on a speaker.

D. It sends data over the serial port.

Solution

1. Step 1: Understand the purpose of tone()

   The tone() function generates a square wave of the specified frequency on a pin, producing sound through a speaker.

2. Step 2: Compare options with function behavior

   Only It plays a sound at a specific frequency on a speaker. describes playing sound at a frequency, which matches tone() usage.

3. Final Answer:

   It plays a sound at a specific frequency on a speaker. -> Option C

4. Quick Check:

   tone() = plays sound [OK]

Hint: Remember: tone() controls sound frequency, not LEDs or sensors [OK]

Common Mistakes:

• Confusing tone() with analogWrite() for LEDs
• Thinking tone() reads sensor data
• Assuming tone() sends serial data

2. Which of the following is the correct syntax to play a 440 Hz tone on pin 8 for 500 milliseconds?

easy

A. tone(8, 440); delay(500);

B. tone(440, 8, 500);

C. tone(8, 500, 440);

D. tone(8, 440, 500);

Solution

1. Step 1: Recall tone() function parameters

   The correct order is tone(pin, frequency, duration). So pin 8, frequency 440 Hz, duration 500 ms.

2. Step 2: Check each option's parameter order

   tone(8, 440, 500); matches the correct order. tone(8, 440); delay(500); plays tone without duration, relying on delay, which is valid but not the asked syntax.

3. Final Answer:

   tone(8, 440, 500); -> Option D

4. Quick Check:

   tone(pin, freq, duration) = tone(8, 440, 500) [OK]

Hint: Remember tone(pin, frequency, duration) order [OK]

Common Mistakes:

• Swapping frequency and pin parameters
• Confusing duration with frequency
• Omitting duration but expecting timed tone

3. What will the following Arduino code do?

void setup() {
  tone(9, 262, 250);
  delay(300);
  tone(9, 294, 250);
  delay(300);
  noTone(9);
}
void loop() {}

medium

A. Play two notes (C4 and D4) on pin 9 with pauses, then stop sound.

B. Play one continuous note at 262 Hz for 800 ms.

C. Play no sound because noTone() stops all tones immediately.

D. Cause a compile error due to missing loop content.

Solution

1. Step 1: Analyze tone() and delay() calls

   The code plays 262 Hz tone for 250 ms, waits 300 ms, then plays 294 Hz tone for 250 ms, waits 300 ms.

2. Step 2: Understand noTone() effect

   noTone(9) stops any tone on pin 9 after the notes play. The loop is empty, so no repetition.

3. Final Answer:

   Play two notes (C4 and D4) on pin 9 with pauses, then stop sound. -> Option A

4. Quick Check:

   tone + delay + noTone = two notes then stop [OK]

Hint: Check tone frequency and delays to predict sound timing [OK]

Common Mistakes:

• Assuming tones play simultaneously
• Thinking noTone() stops before notes play
• Believing empty loop causes error

4. Which statement correctly describes the behavior of this code snippet:

void setup() {
  tone(7, 330);
  delay(500);
  tone(7, 349);
  delay(500);
  noTone(7);
}
void loop() {}

medium

A. The delay() calls are too short to hear the notes.

B. The second tone replaces the first (which continues until replaced), resulting in sequential notes without overlap.

C. noTone() is called too early, stopping the melody immediately.

D. tone() requires a duration parameter to work correctly.

Solution

1. Step 1: Understand tone() without duration

   Calling tone(pin, freq) plays the tone continuously until stopped or replaced by another tone() on the same pin.

2. Step 2: Analyze timing of tones and delays

   The first tone starts and plays for 500 ms during delay(). The second tone replaces it immediately, playing the new frequency for the next 500 ms. No overlap occurs.

3. Final Answer:

   The second tone replaces the first (which continues until replaced), resulting in sequential notes without overlap. -> Option B

4. Quick Check:

   tone without duration replaced by next tone() = sequential notes [OK]

Hint: tone() without duration continues until next tone() or noTone() [OK]

Common Mistakes:

• Assuming tone() stops automatically after delay
• Thinking tone() calls on same pin cause overlap
• Believing delay controls tone length without duration

5. You want to play a melody of 3 notes (262 Hz, 294 Hz, 330 Hz) each lasting 300 ms on pin 10 with 100 ms pause between notes. Which code snippet correctly achieves this?

hard

A. for (int i = 0; i < 3; i++) {
int notes[] = {262, 294, 330};
tone(10, notes[i], 300);
delay(400);
}
noTone(10);

B. int notes[] = {262, 294, 330};
for (int i = 0; i < 3; i++) {
tone(10, notes[i]);
delay(250);
noTone(10);
delay(100);
}

C. int notes[] = {262, 294, 330};
for (int i = 0; i < 3; i++) {
tone(10, notes[i], 300);
delay(250);
noTone(10);
delay(100);
}

D. int notes[] = {262, 294, 330};
for (int i = 0; i < 3; i++) {
tone(10, notes[i]);
delay(400);
}
noTone(10);

Solution

1. Step 1: Understand timing requirements

   Each note plays 300 ms, plus 100 ms pause, total 400 ms delay between notes.

2. Step 2: Check code for correct tone duration and delay

   for (int i = 0; i < 3; i++) { int notes[] = {262, 294, 330}; tone(10, notes[i], 300); delay(400); } noTone(10); uses tone with duration 300 ms and delay 400 ms, matching timing. noTone(10) after loop stops sound.

3. Step 3: Verify other options

   int notes[] = {262, 294, 330}; for (int i = 0; i < 3; i++) { tone(10, notes[i]); delay(250); noTone(10); delay(100); } calls noTone inside loop, cutting notes short. int notes[] = {262, 294, 330}; for (int i = 0; i < 3; i++) { tone(10, notes[i], 300); delay(250); noTone(10); delay(100); } delays only 250 ms, missing pause. int notes[] = {262, 294, 330}; for (int i = 0; i < 3; i++) { tone(10, notes[i]); delay(400); } noTone(10); plays tone without duration, causing overlap.

4. Final Answer:

   for (int i = 0; i < 3; i++) { int notes[] = {262, 294, 330}; tone(10, notes[i], 300); delay(400); } noTone(10); -> Option A

5. Quick Check:

   tone(duration) + delay(total note+pause) = for (int i = 0; i < 3; i++) { int notes[] = {262, 294, 330}; tone(10, notes[i], 300); delay(400); } noTone(10); [OK]

Hint: Use tone(duration) plus delay(total note+pause) for melody timing [OK]

Common Mistakes:

• Calling noTone() inside loop too early
• Using delay shorter than note+pause
• Omitting duration causing overlapping notes