Sound Physics Problems With Solutions For Grade 11

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Let's dive into some sound physics problems perfect for you 11th graders! Physics can be a tough subject, especially when we're talking about tricky concepts like sound. But don't worry, guys! We're gonna break it down with clear explanations and step-by-step solutions. By the end of this, you’ll be tackling sound problems like a pro. So buckle up, grab your calculators, and let’s get started!

Understanding Sound Waves

Before we jump into solving problems, it's super important to understand the basics of sound waves. Sound waves are longitudinal waves, meaning that the particles in the medium (like air, water, or solids) vibrate parallel to the direction the wave is traveling. Think of it like a slinky: when you push one end, the compression travels along the slinky. That compression and rarefaction (expansion) is how sound travels. Key characteristics of sound waves include:

  • Frequency (f): This is the number of complete waves that pass a point in one second, measured in Hertz (Hz). Frequency determines the pitch of the sound; a higher frequency means a higher pitch.
  • Wavelength (λ): This is the distance between two consecutive crests or troughs of a wave, usually measured in meters (m). Wavelength is related to frequency and speed of sound.
  • Speed (v): This is how fast the sound wave travels through the medium, usually measured in meters per second (m/s). The speed of sound depends on the medium; it's faster in solids and liquids than in gases.
  • Amplitude: This is the maximum displacement of the particles from their resting position. Amplitude determines the loudness or intensity of the sound; a larger amplitude means a louder sound.

The relationship between these characteristics is given by the wave equation:

v = fλ

Where:

  • v is the speed of sound
  • f is the frequency
  • λ is the wavelength

Understanding these fundamentals is crucial because they form the basis for solving more complex problems. You'll often need to manipulate this equation to find unknown values. So, make sure you're comfortable with it. Also, keep in mind that the speed of sound in air at room temperature (around 20°C) is approximately 343 m/s. This value can change with temperature, so pay close attention to the information provided in the problem statement.

Sample Problems and Solutions

Okay, let's get to the fun part: solving problems! I'll give you a few examples of the types of sound physics problems you might encounter. I will show you how to approach them. Remember, the key is to break down each problem into smaller, more manageable steps.

Problem 1: Wavelength Calculation

Problem: A tuning fork vibrates at a frequency of 440 Hz. If the speed of sound in air is 343 m/s, what is the wavelength of the sound wave produced?

Solution:

  1. Identify the knowns:
    • Frequency (f) = 440 Hz
    • Speed of sound (v) = 343 m/s
  2. Identify the unknown:
    • Wavelength (λ) = ?
  3. Use the wave equation:
    • v = fλ
  4. Rearrange the equation to solve for wavelength:
    • λ = v / f
  5. Plug in the values and calculate:
    • λ = 343 m/s / 440 Hz
    • λ ≈ 0.78 m

Answer: The wavelength of the sound wave is approximately 0.78 meters.

Problem 2: Frequency Calculation

Problem: A sound wave has a wavelength of 2.5 meters and travels at a speed of 345 m/s. What is the frequency of the sound wave?

Solution:

  1. Identify the knowns:
    • Wavelength (λ) = 2.5 m
    • Speed of sound (v) = 345 m/s
  2. Identify the unknown:
    • Frequency (f) = ?
  3. Use the wave equation:
    • v = fλ
  4. Rearrange the equation to solve for frequency:
    • f = v / λ
  5. Plug in the values and calculate:
    • f = 345 m/s / 2.5 m
    • f = 138 Hz

Answer: The frequency of the sound wave is 138 Hz.

Problem 3: Speed of Sound Calculation

Problem: A sound wave with a frequency of 500 Hz has a wavelength of 0.686 meters. What is the speed of the sound wave?

Solution:

  1. Identify the knowns:
    • Frequency (f) = 500 Hz
    • Wavelength (λ) = 0.686 m
  2. Identify the unknown:
    • Speed of sound (v) = ?
  3. Use the wave equation:
    • v = fλ
  4. Plug in the values and calculate:
    • v = 500 Hz * 0.686 m
    • v = 343 m/s

Answer: The speed of the sound wave is 343 m/s.

Intensity and Loudness

Besides frequency, wavelength, and speed, two other crucial concepts are intensity and loudness. The intensity of a sound wave is the power per unit area carried by the wave, usually measured in watts per square meter (W/m²). It’s directly proportional to the square of the amplitude of the sound wave. Loudness, on the other hand, is a subjective perception of sound intensity by the human ear. It's often measured in decibels (dB).

The relationship between intensity (I) and loudness (β) in decibels is given by:

β = 10 log₁₀ (I / I₀)

Where:

  • β is the loudness in decibels (dB)
  • I is the intensity of the sound wave
  • I₀ is the reference intensity, which is the threshold of hearing (1.0 x 10⁻¹² W/m²)

Problem 4: Intensity Calculation

Problem: The loudness of a sound is measured to be 60 dB. What is the intensity of the sound wave?

Solution:

  1. Identify the knowns:
    • Loudness (β) = 60 dB
    • Reference intensity (I₀) = 1.0 x 10⁻¹² W/m²
  2. Identify the unknown:
    • Intensity (I) = ?
  3. Use the formula for loudness:
    • β = 10 log₁₀ (I / I₀)
  4. Rearrange the equation to solve for intensity:
    • 60 = 10 log₁₀ (I / 1.0 x 10⁻¹²)
    • 6 = log₁₀ (I / 1.0 x 10⁻¹²)
    • 10⁶ = I / 1.0 x 10⁻¹²
    • I = 10⁶ * 1.0 x 10⁻¹²
    • I = 1.0 x 10⁻⁶ W/m²

Answer: The intensity of the sound wave is 1.0 x 10⁻⁶ W/m².

Problem 5: Loudness Calculation

Problem: The intensity of a sound wave is 1.0 x 10⁻⁵ W/m². What is the loudness of the sound in decibels?

Solution:

  1. Identify the knowns:
    • Intensity (I) = 1.0 x 10⁻⁵ W/m²
    • Reference intensity (I₀) = 1.0 x 10⁻¹² W/m²
  2. Identify the unknown:
    • Loudness (β) = ?
  3. Use the formula for loudness:
    • β = 10 log₁₀ (I / I₀)
  4. Plug in the values and calculate:
    • β = 10 log₁₀ (1.0 x 10⁻⁵ W/m² / 1.0 x 10⁻¹² W/m²)
    • β = 10 log₁₀ (10⁷)
    • β = 10 * 7
    • β = 70 dB

Answer: The loudness of the sound is 70 dB.

Doppler Effect

The Doppler effect is the change in frequency of a sound wave due to the motion of the source, the observer, or both. When a sound source moves towards you, the frequency appears higher (higher pitch), and when it moves away, the frequency appears lower (lower pitch). The formula for the observed frequency (f') is:

f' = f (v ± v₀) / (v ± vₛ)

Where:

  • f' is the observed frequency
  • f is the source frequency
  • v is the speed of sound in the medium
  • v₀ is the speed of the observer
  • vₛ is the speed of the source

Important:

  • Use + v₀ when the observer is moving towards the source.
  • Use - v₀ when the observer is moving away from the source.
  • Use - vₛ when the source is moving towards the observer.
  • Use + vₛ when the source is moving away from the observer.

Problem 6: Doppler Effect Calculation (Moving Source)

Problem: A car is moving towards you at a speed of 20 m/s, honking its horn, which has a frequency of 500 Hz. If the speed of sound is 343 m/s, what frequency do you hear?

Solution:

  1. Identify the knowns:
    • Source frequency (f) = 500 Hz
    • Speed of sound (v) = 343 m/s
    • Speed of source (vₛ) = 20 m/s (moving towards)
    • Speed of observer (v₀) = 0 m/s (stationary)
  2. Identify the unknown:
    • Observed frequency (f') = ?
  3. Use the Doppler effect formula:
    • f' = f (v ± v₀) / (v ± vₛ)
  4. Plug in the values and calculate:
    • Since the source is moving towards the observer, we use - vₛ in the denominator.
    • f' = 500 Hz * (343 m/s + 0 m/s) / (343 m/s - 20 m/s)
    • f' = 500 Hz * (343 m/s) / (323 m/s)
    • f' ≈ 531 Hz

Answer: You hear a frequency of approximately 531 Hz.

Problem 7: Doppler Effect Calculation (Moving Observer)

Problem: A train is stationary, sounding its whistle at a frequency of 400 Hz. You are standing on a platform, and the train starts moving away from you at 15 m/s. The speed of sound is 343 m/s. What frequency do you observe as the train moves away?

Solution:

  1. Identify the knowns:
    • Source frequency (f) = 400 Hz
    • Speed of sound (v) = 343 m/s
    • Speed of source (vₛ) = 0 m/s (stationary initially)
    • Speed of observer (v₀) = 15 m/s (moving away)
  2. Identify the unknown:
    • Observed frequency (f') = ?
  3. Use the Doppler effect formula:
    • f' = f (v ± v₀) / (v ± vₛ)
    • Since the observer is moving away from the source and the source is stationary, we adjust the formula.
    • f' = f (v - v₀) / v
  4. Plug in the values and calculate:
    • f' = 400 Hz * (343 m/s - 15 m/s) / 343 m/s
    • f' = 400 Hz * (328 m/s) / 343 m/s
    • f' ≈ 382 Hz

Answer: You observe a frequency of approximately 382 Hz as the train moves away.

Practice Makes Perfect

Alright, guys, we've covered quite a bit about sound physics problems! Remember, the key to mastering these concepts is practice. Work through as many problems as you can find, and don't be afraid to ask for help when you get stuck. Understanding the underlying principles and knowing how to apply the formulas is essential. Physics can be challenging, but with dedication and consistent effort, you'll be acing those sound problems in no time! Keep practicing, stay curious, and you'll conquer the world of physics! Good luck!