Problem 70 - Pearson Physics

Question

A tuning fork with a frequency of \(320.0 \mathrm{~Hz}\) and a tuning fork with an unknown frequency produce beats with a frequency of \(4.5 \mathrm{~Hz}\). If the frequency of the \(320.0-\mathrm{Hz}\) fork is lowered slightly by placing a bit of putty on one of its tines, the new beat frequency is \(7.5 \mathrm{~Hz}\). (a) Which tuning fork has the lower frequency? Explain. (b) What is the final frequency of the \(320.0\)-Hz tuning fork? (c) What is the frequency of the other tuning fork?

Step-by-Step Solution

Answer

(a) 320 Hz fork is lower. (b) 317.0 Hz. (c) 324.5 Hz.

Step 1: Understanding Beat Frequency

The beat frequency is the absolute difference between two frequencies. Given: \[ f_1 = 320.0 \text{ Hz (known tuning fork)} \]\[ \text{Beat frequency} = 4.5 \text{ Hz} \]Therefore, the possible frequencies of the unknown tuning fork \( f_2 \) can be determined by:\[ f_2 = f_1 \pm \text{Beat frequency} \]Which gives us two possibilities:\[ f_2 = 320.0 + 4.5 = 324.5 \text{ Hz} \]\[ f_2 = 320.0 - 4.5 = 315.5 \text{ Hz} \]

Step 2: Evaluating the Tuning Forks after the Change

When putty is added to the 320.0 Hz fork, its frequency lowers, causing a change in the beat frequency. The new beat frequency is 7.5 Hz. Thus, the new frequency of the altered fork, \( f'_1 \), creates the new beat:\[ |f'_1 - f_2| = 7.5 \text{ Hz} \]

Step 3: Applying the Beat Frequency Change

Determine the new frequency of the altered fork through two cases from Step 1:- If the unknown fork initially was 324.5 Hz, and now the new beat frequency is 7.5 Hz after reducing the frequency of the 320 Hz fork, then:\[ 324.5 - f'_1 = 7.5 \Rightarrow f'_1 = 324.5 - 7.5 = 317.0 \text{ Hz} \]- If it was initially 315.5 Hz, then:\[ f'_1 - 315.5 = 7.5 \Rightarrow f'_1 = 315.5 + 7.5 = 323.0 \text{ Hz} \]

Step 4: Concluding the Correct Frequency for the Unknown Fork

Cross-validation helps confirm the initially correct assumption. Assuming a frequency, we align the resultant beat frequency with lower frequency changes. The initial fork must have been 324.5 Hz because the 320.0 Hz shifted closer to it (317.0 Hz) resulting in a new beat frequency of 7.5 Hz.

Step 5: Answering the Questions

(a) The unknown fork was initially 324.5 Hz, which is higher than the 320 Hz fork. (b) The final frequency of the 320 Hz fork is 317.0 Hz after adding the putty. (c) The frequency of the other fork is 324.5 Hz.

Key Concepts

Tuning Forks

A tuning fork is a small instrument usually made of steel or aluminum. It consists of a handle and two tines that vibrate when struck. These vibrations create a pure tone used as a standard for tuning musical instruments.

• Tuning forks produce sound waves at a specific frequency.
• The vibration of the fork is quite consistent, which makes it reliable for experimentation.
• When two tuning forks are used together, differences in their frequencies can be identified through beat frequencies.

Understanding tuning forks is crucial for solving sound-related problems because they are precise and predictable in their frequency output. They serve as a perfect baseline for comparison when examining how sound waves interact with one another.

Frequency Calculation

Frequency calculation involves determining the number of wave cycles that occur in a second, measured in hertz (Hz). The exercise revolves around calculating the unknown frequency using the concept of beat frequencies.
To solve such exercises, follow these steps:

• Identify the known frequency of one fork. In this problem, it's 320 Hz.
• Determine the beat frequency, which is the difference between the frequencies of the two forks. Here, two beat frequencies are given: 4.5 Hz initially and 7.5 Hz after the frequency change.
• Calculate possible frequencies for the unknown fork using the formula: \( f_2 = f_1 \pm ext{Beat frequency} \). This yields two potential frequencies. For example, when the known frequency is 320 Hz and the beat frequency is 4.5 Hz, the possible unknown frequencies are 324.5 Hz or 315.5 Hz.

This kind of calculation helps in accurately identifying the characteristics of sound waves when given partial data. Mathematically deducing the unknown quantities from known parameters aids in grasping how sound manipulation affects wave interactions.

Wave Interference

Wave interference occurs when two sound waves meet. They can either amplify each other (constructive interference) or reduce each other (destructive interference), resulting in various sound phenomena.
Beat frequencies are a type of interference. They are the result of closely matched frequencies causing fluctuations in loudness as the waves alternately reinforce and cancel each other. Here’s how it works:

• If two waves have slightly different frequencies, the interference causes a modulation in amplitude.
• This modulation is heard as beats, which are perceived as a wobbling sound intensity.
• The frequency of the beats is the absolute difference between the frequencies of the two waves. For instance, if you have a 320 Hz fork and one of possibly 324.5 Hz or 315.5 Hz, the beats are at either 4.5 Hz or 7.5 Hz, respectively.

Through understanding wave interference, we gain insight into the behavior of sound in various environments and how changes in conditions can affect our auditory perceptions.