To solve the problem of determining the frequency of the loaded tuning fork, we need to understand the concept of beats and how they relate to the frequencies of the two tuning forks. When two tuning forks are sounded together, the beats produced are a result of the interference between the sound waves from each fork. The number of beats per second is equal to the absolute difference in their frequencies.
Understanding Beats
When two sound waves of slightly different frequencies interact, they create a phenomenon known as beats. The formula for the number of beats per second (B) is given by:
B = |f1 - f2|
where f1 and f2 are the frequencies of the two tuning forks. In this case, we know that:
- The frequency of the first tuning fork (f1) is 368 Hz.
- When the second fork is loaded with wax, it produces 20 beats in 4 seconds.
Calculating the Frequency of the Loaded Fork
First, we need to determine the number of beats per second produced by the loaded fork:
Beats per second = Total beats / Time
Here, we have:
- Total beats = 20
- Time = 4 seconds
Calculating this gives:
Beats per second = 20 beats / 4 seconds = 5 beats per second
Now, we can use the beats formula to find the frequency of the loaded fork (f2). Since we know the number of beats per second is 5, we can set up the equation:
5 = |368 Hz - f2|
Solving for f2
This absolute value equation can yield two possible scenarios:
- 368 Hz - f2 = 5
- f2 - 368 Hz = 5
Let's solve each case:
Case 1: 368 Hz - f2 = 5
Rearranging gives:
f2 = 368 Hz - 5 = 363 Hz
Case 2: f2 - 368 Hz = 5
Rearranging gives:
f2 = 368 Hz + 5 = 373 Hz
Final Frequencies
Thus, the frequency of the loaded tuning fork can either be 363 Hz or 373 Hz. To determine which one is correct, we consider that loading the fork with wax typically lowers its frequency. Therefore, the more reasonable answer is:
f2 = 363 Hz
Summary
In summary, by understanding the relationship between the frequencies of the two tuning forks and the beats produced, we calculated that the frequency of the loaded tuning fork is 363 Hz. This example illustrates how sound waves interact and how we can use mathematical relationships to solve real-world problems in acoustics.
