Problem 64 - Pearson Physics

Question

Predict \& Explain (a) Is the beat frequency produced when a \(245-\mathrm{Hz}\) tone and a \(240-\mathrm{Hz}\) tone are played together greater than, less than, or equal to the beat frequency produced when a \(140-\mathrm{Hz}\) tone and a \(145-\mathrm{Hz}\) tone are played together? (b) Choose the best explanation from among the following: A. The beat frequency is determined by the difference between the frequencies and is independent of their actual values. B. The higher frequencies will produce a greater beat frequency. C. The percent change in frequency for \(240 \mathrm{~Hz}\) and \(245 \mathrm{~Hz}\) is less than for \(140 \mathrm{~Hz}\) and \(145 \mathrm{~Hz}\), resulting in a smaller beat frequency.

Step-by-Step Solution

Answer

The beat frequencies are equal; best explanation: A.

Step 1: Calculate Beat Frequency for 245 Hz and 240 Hz

The formula for beat frequency is the absolute difference between the two frequencies. Calculate the beat frequency for the tones with frequencies 245 Hz and 240 Hz using the formula: \[ f_b = |245 - 240| = 5 \text{ Hz} \]

Step 2: Calculate Beat Frequency for 140 Hz and 145 Hz

Use the same formula for beat frequency: \[ f_b = |145 - 140| = 5 \text{ Hz} \]

Step 3: Compare Beat Frequencies

The beat frequency for the first pair is 5 Hz and for the second pair is also 5 Hz. Thus, the beat frequency is equal for both pairs of tones.

Step 4: Determine the Best Explanation

Option A states that the beat frequency is determined by the difference between frequencies and is independent of their actual values. Since both beat frequencies calculated are equal, based solely on their differences, Option A is the best explanation.

Key Concepts

Difference Between Frequencies

Beat frequency arises when two waves of nearby but different frequencies overlap. The sensation you hear as a beat is due to the interference between these frequencies. Importantly, the beat frequency is determined by the difference between these frequencies. Imagine you have two sounds: one at \(245 \text{ Hz}\) and another at \(240 \text{ Hz}\). The beat frequency would be the absolute value of the difference:

• \(f_b = |245 - 240| = 5 \text{ Hz}\)

This tells us that even though you have specific tones, what matters for the beat frequency is how far apart they are, frequency-wise. Now, let's consider two other tones at \(140 \text{ Hz}\) and \(145 \text{ Hz}\). The calculation is similar:

• \(f_b = |145 - 140| = 5 \text{ Hz}\)

In both cases, the difference is 5 Hz, leading to the same beat frequency. So, no matter the actual frequency values, just the difference matters for the beat frequency when two waves combine.

Independent of Actual Frequency Values

A remarkable aspect of beat frequencies is their independence from the actual values of the frequencies involved. This means that whether you are dealing with a much higher or lower pair of frequencies, the beat frequency calculation focuses solely on the difference. For instance:

• A pair of waves at \(245 \text{ Hz}\) and \(240 \text{ Hz}\)
• An entirely different set of \(400 \text{ Hz}\) and \(395 \text{ Hz}\)

Both would result in the same beat frequency of 5 Hz, illustrating beautifully this concept of independence. This characteristic is particularly important for scenarios like tuning instruments, where musicians listen for beats to achieve perfect harmony. However, the actual frequency numbers don't affect the beat frequency, only their difference does.

Formula for Beat Frequency

Understanding the formula for beat frequency is crucial for grasping how these audio phenomena occur. The fundamental formula is:

• \[ f_b = |f_1 - f_2| \]

Here, \( f_1 \) and \( f_2 \) are the frequencies of the two tones you are comparing. The formula simply calls for taking the absolute difference between these two frequencies, making it straightforward and universally applicable. For example, if you are dealing with frequencies of \(245 \text{ Hz}\) and \(240 \text{ Hz}\), the formula computes:

• \[ f_b = |245 - 240| = 5 \text{ Hz} \]

So, no matter the complexity of the frequencies involved, this simple subtraction gives you the beat frequency. This formula elegantly captures the essence of the phenomenon, offering a reliable means to predict and explain how two different sound waves will interact in your ears.