Problem 22 - Physics

Question

Two tuning forks \(X\) and \(Y\) have different frequencies and produce an \(8-H z\) beat frequency when sounded together. When \(X\) is sounded along with a \(392-\mathrm{Hz}\) tone, a \(3-\mathrm{Hz}\) beat frequency is detected. When \(Y\) is sounded along with the \(392-\mathrm{Hz}\) tone, a \(5-\mathrm{Hz}\) beat frequency is heard. What are the frequencies \(f_{X}\) and \(f_{Y}\) when \((a) f_{X}\) is greater than \(f_{Y}\) and (b) \(f_{X}\) is less than \(f_{Y}\) ?

Step-by-Step Solution

Answer

(a) \( f_X = 395 \text{ Hz} \), \( f_Y = 387 \text{ Hz} \); (b) \( f_X = 389 \text{ Hz} \), \( f_Y = 397 \text{ Hz} \).

Step 1: Understanding the Beat Frequency

The problem involves beat frequencies. Two sound sources of frequencies \( f_X \) and \( f_Y \) together produce a beat frequency of 8 Hz. This means the difference between the two forks' frequencies is \( |f_X - f_Y| = 8 \text{ Hz} \).

Step 2: Use X and 392 Hz Tone Relation

When fork \( X \) is played with a 392 Hz tone, a beat frequency of 3 Hz is detected. This tells us that \( |f_X - 392| = 3 \), so the frequency \( f_X \) can either be 395 Hz or 389 Hz.

Step 3: Use Y and 392 Hz Tone Relation

When fork \( Y \) is played with a 392 Hz tone, a beat frequency of 5 Hz is detected. This implies that \( |f_Y - 392| = 5 \), so the frequency \( f_Y \) could be either 397 Hz or 387 Hz.

Step 4: Determine Frequencies for Case (a)

For \( f_X > f_Y \), let's consider \( f_X = 395 \text{ Hz} \) and \( f_Y = 387 \text{ Hz} \) because the difference, 8 Hz, matches the condition \( f_X - f_Y = 8 \).

Step 5: Determine Frequencies for Case (b)

When \( f_X < f_Y \), we select \( f_X = 389 \text{ Hz} \) and \( f_Y = 397 \text{ Hz} \), resulting in a difference that satisfies the equation \( f_Y - f_X = 8 \).

Step 6: State the Results

In conclusion, for (a) \( f_X = 395 \text{ Hz} \) and \( f_Y = 387 \text{ Hz} \); for (b) \( f_X = 389 \text{ Hz} \) and \( f_Y = 397 \text{ Hz} \).

Key Concepts

Tuning Forks

Tuning forks are simple yet fascinating tools that help us understand the concept of sound frequencies. They consist of a U-shaped metal bar that vibrates at a specific frequency when struck.
These vibrations produce sound waves in the air, which we perceive as musical notes. Tuning forks are commonly used in musical instrument tuning and physics experiments. They provide a clear and steady tone that is helpful in measuring other frequencies, like those used in the problem above.
When two or more tuning forks are used together, they can produce a phenomenon known as "beats," which enables us to determine differences in frequency.

Sound Waves

Sound waves are vibrations that travel through a medium like air, water, or solids. They are created by vibrating objects, such as tuning forks, which push and pull the molecules in the air.
This movement causes alternating areas of high and low pressure, known as compressions and rarefactions. The frequency of a sound wave, measured in Hertz (Hz), indicates how many compressions pass a point per second. Higher frequencies mean more cycles per second, producing higher-pitched sounds, whereas lower frequencies give lower-pitched sounds.

• Sound waves can interfere with each other. When two sound waves of different frequencies are played together, they can produce a new wave pattern called "beats," as seen in the beat frequency problem.
• These beats are a result of constructive and destructive interference, causing the sound to alternately increase and decrease in intensity.

Understanding sound wave properties is crucial in solving frequency-related problems in physics.

Frequency Difference

The frequency difference refers to the variation in the number of oscillations per second between two sound waves. It is essential in determining beat frequency.
Beat frequency equals the absolute difference in frequencies for two waves, represented mathematically as \(|f_X - f_Y|\). In the context of tuning forks, the beat frequency is a significant indicator of how close or apart the frequencies are.

• When the frequencies are very close, the beat frequency is low, producing a slow wavering sound.
• Conversely, when they are farther apart, the beat frequency is higher, resulting in rapid fluctuations.

Understanding this concept allows us to solve problems where the precise calculation of individual frequencies affects results, like in tuning musical instruments or scientific experiments.

Physics Problem Solving

Solving physics problems often involves breaking down complex phenomena into simpler, understandable parts. Here’s a structured approach to solving frequency-related problems, like the tuning fork scenario provided:

• Identify the Given Information: Recognize all the frequencies and beat frequencies mentioned in the problem. For instance, the provided values were 8 Hz, 3 Hz, and 5 Hz.
• Understand the Concept: Familiarize yourself with concepts like beat frequency and how sound waves interact. This understanding is crucial to forming equations that represent the situation.
• Form Equations: Using the knowledge of beat frequency, set up mathematical expressions. For example, knowing that \(|f_X - f_Y| = 8\) helps begin the problem-solving process.
• Solve Step-by-Step: Each equation formed is solved individually, reducing the complexity. Evaluate each condition separately, such as when \(f_X > f_Y\) or \(f_X < f_Y\).
• Check Results: Ensure the solutions satisfy all conditions given in the original problem.

Approaching problems in this organized manner makes it easier to solve physics exercises efficiently and accurately.