Problem 40 - Elementary and Intermediate Algebra

Question

Solve each problem by writing a variation model. Organ Pipes. The frequency of vibration of air in an organ pipe is inversely proportional to the length of the pipe. If a pipe 2 feet long vibrates 256 times per second, how many times per second will a 6 -foot pipe vibrate? CAN'T COPY THE IMAGE

Step-by-Step Solution

Answer

A 6-foot pipe will vibrate approximately 85.33 times per second.

Step 1: Understand the Relationship

The problem states that the frequency of vibration is inversely proportional to the length of the pipe. This can be mathematically represented as \( f \propto \frac{1}{L} \), where \( f \) is the frequency and \( L \) is the length of the pipe.

Step 2: Write the Equation for Inverse Proportion

The equation for inverse proportionality can be expressed as \( f = \frac{k}{L} \), where \( k \) is a constant of proportionality. We need to find \( k \) using the given information.

Step 3: Find the Constant of Proportionality

Utilize the information that a 2-foot pipe has a frequency of 256 Hz. Substitute these into the equation: \( 256 = \frac{k}{2} \). Solve for \( k \) by multiplying both sides by 2: \( k = 256 \times 2 = 512 \).

Step 4: Calculate the Frequency for a 6-foot Pipe

Now that we have \( k = 512 \), substitute \( k \) and \( L = 6 \) feet into the equation \( f = \frac{k}{L} \) to find the new frequency: \( f = \frac{512}{6} \approx 85.33 \).

Step 5: Conclusion

The frequency of vibration for a 6-foot pipe is approximately 85.33 times per second (rounded to two decimal places).

Key Concepts

Frequency and Length Relationship

Understanding the relationship between frequency and length is crucial when studying the behavior of sound in an organ pipe. In this context, frequency refers to how many times the air inside the pipe vibrates in a second, measured in Hertz (Hz). Length of the pipe is simply the measurement from one end of the pipe to the other, usually in feet or meters.

In the given problem, the frequency of vibration is inversely proportional to the length of the pipe. This simply means that as one value increases, the other decreases. Mathematically, this inverse relationship is expressed as:

• \[ f \propto \frac{1}{L} \]

Here, \( f \) represents frequency, and \( L \) stands for the length of the pipe.

This kind of relationship tells us that a shorter pipe will have a higher frequency, making a higher pitched sound, while a longer pipe will vibrate less frequently, producing a lower pitched sound. In simple terms, the longer the pipe, the slower the vibrations, and thus, lower the frequency.

Constant of Proportionality

In the equation describing inverse proportionality, the constant of proportionality, often denoted as \( k \), is a key element. This constant links the frequency and length in a specific relationship for the particular organ pipe.

To find the constant \( k \), we take a known pair of frequency and length values. In this example, we know a 2-foot pipe vibrates at 256 Hz, represented as:

• \[ f = \frac{k}{L} \]

Substituting the knowns: \( 256 = \frac{k}{2} \). Solving for \( k \), multiply both sides by 2:

• \[ k = 256 \times 2 = 512 \]

The constant \( k = 512 \) belongs to this particular set of conditions and helps us find the frequency for other lengths of pipe. In simpler terms, the constant of proportionality acts like a bridge or translator, allowing us to switch between frequency and pipe length within a specific context.

Solving Equations with Variation

Solving equations that involve inverse variation can seem complex at first, but it's a systematic process. Once we have determined the constant of proportionality, \( k \), solving for different variables becomes straightforward.

In our example, with \( k = 512 \) and knowing the inverse relationship formula, you can find out how many times per second a 6-foot pipe will vibrate. Use the equation:

• \[ f = \frac{k}{L} \]

Substitute \( k = 512 \) and \( L = 6 \):

• \[ f = \frac{512}{6} \approx 85.33 \]

So, a 6-foot organ pipe vibrates approximately 85.33 times per second. This step-by-step problem-solving approach can be used for any inverse variation situation, where you determine the constant first and then use it to find unknown values.