Problem 14 - Pearson Physics

Question

The frequency of a sound is doubled. Does the wavelength of this sound increase, decrease, or stay the same? Explain.

Step-by-Step Solution

Answer

The wavelength decreases when the frequency is doubled.

Step 1: Understanding the Relationship

Sound waves, like any wave, have a specific relationship between their frequency, wavelength, and velocity. The formula that connects these quantities is given by: \[ v = f \lambda \]Here, \( v \) represents the velocity of sound, \( f \) is the frequency, and \( \lambda \) is the wavelength.

Step 2: Identifying Constants

In this problem, the velocity of sound \( v \) is constant because it depends on the medium through which the sound is traveling — which we assume doesn't change. Therefore, any change in frequency must result in an opposite change in wavelength to keep the velocity unchanged.

Step 3: Applying Changes to the Formula

According to the problem, the frequency \( f \) is doubled. Let's denote the original frequency as \( f_0 \) and the original wavelength as \( \lambda_0 \). After doubling the frequency, the new frequency is \( 2f_0 \).

Step 4: Rearranging the Formula

Using the formula \( v = f \lambda \), and substituting the doubled frequency, we get:\[ v = 2f_0 \lambda' \]Where \( \lambda' \) is the new wavelength. Since \( v = f_0 \lambda_0 \) initially, we equate both expressions to find:\[ f_0 \lambda_0 = 2f_0 \lambda' \]

Step 5: Solving for the New Wavelength

Cancel out the frequency \( f_0 \) from both sides:\[ \lambda_0 = 2\lambda' \]Now, solve for \( \lambda' \):\[ \lambda' = \frac{\lambda_0}{2} \]Thus, the new wavelength is half of the original wavelength.

Key Concepts

Wavelength

The term "wavelength" refers to the distance between successive peaks of a sound wave, which is crucial in determining the characteristics of the sound. Wavelength is denoted by the Greek letter \( \lambda \) and is measured in meters. In sound waves, wavelength is inversely proportional to frequency when the velocity of sound is constant. This means that if you increase the frequency, the wavelength decreases and vice versa, given that the sound travels through the same medium.
For example, when the frequency doubles, the wavelength becomes half of its original value. This relationship can be understood using the core equation \( v = f \lambda \), where each component plays a vital role in the wave's behavior within a given medium.

Frequency

Frequency measures how often the sound wave oscillates or vibrates per second, and it is measured in Hertz (Hz). High frequency means more oscillations per second, leading to a higher pitch sound, whereas lower frequency results in a lower pitched sound. When the frequency of a sound wave increases, assuming the medium through which the sound is traveling remains unchanged, the wavelength will decrease to maintain constant velocity.
Using our exercise's context, the frequency was doubled, showing how flexible and inverse its relation is with wavelength. The fact that a higher frequency results in a shorter wavelength aligns with how sound behaves in mediums like air.

Velocity of Sound

The velocity of sound is the speed at which sound waves travel through a medium. This property is denoted by \( v \) and is typically measured in meters per second (m/s). The velocity depends on factors like the medium's density and temperature, but it remains constant as long as these factors don't change.
When evaluating sound waves in consistent mediums, the velocity helps maintain a balance between frequency and wavelength. In our original problem, the velocity remains constant, ensuring that any change in frequency demands a proportional inverse change in wavelength, in order not to disturb the sound wave's propagation speed through the medium.