Question

The period for a certain wave is 34 ms. If there is a Doppler shift that doubles the perceived frequency, which of the following must be true? The detector is moving toward the source at a velocity equal to the speed of sound. The source is moving toward the detector at a velocity equal to half the speed of sound. The perceived period is \(17 \mathrm{ms}\) The perceived period is \(68 \mathrm{ms}\) (A) III only (B) \(\quad\) I and IV only (C) II and III only (D) \(\quad\) I \(,\) II, and IV only

Short answer

A

Step-by-step solution

Understand the given period and frequency

The period of a wave is the reciprocal of the frequency. Given the period is 34 milliseconds (ms), the frequency is calculated as follows:\[ f = \frac{1}{T} \]Where \( f \) is the frequency and \( T \) is the period. Substituting the given period:\[ f = \frac{1}{34 \times 10^{-3} \text{ s}} = 29.41 \text{ Hz} \]

Determine the new frequency after Doppler shift

The problem states that there is a Doppler shift that doubles the perceived frequency. Hence, the new perceived frequency \( f' \) is:\[ f' = 2f = 2 \times 29.41 \text{ Hz} = 58.82 \text{ Hz} \]

Calculate the new perceived period

Using the new perceived frequency, the new period \( T' \) is:\[ T' = \frac{1}{f'} = \frac{1}{58.82 \text{ Hz}} \]Thus:\[ T' = 17 \text{ ms} \]

Analyze the given choices

Let's analyze the given multiple-choice options:1. The detector is moving toward the source at a velocity equal to the speed of sound.2. The source is moving toward the detector at a velocity equal to half the speed of sound.3. The perceived period is 17 ms.4. The perceived period is 68 ms.Based on our calculations, the perceived period due to the Doppler shift is 17 ms.

Choose the correct answer

From the given options:(A) III only(B) I and IV only(C) II and III only(D) I, II, and IV onlyThe perceived period being 17 ms is the only correct statement from our analysis. Therefore, the correct answer is:(A) III only.

Key concepts

wave period calculation

The period of a wave is the time it takes for one complete cycle to pass a given point. It is a fundamental concept in wave mechanics that helps us understand the behavior of various types of waves. The formula to calculate the period (\( T \)) of a wave when the frequency (\( f \)) is known is given by:
\( T = \frac{1}{f} \)
In the example exercise, we start knowing the period of a wave is 34 milliseconds (ms). To find the frequency, we use the reciprocal relationship between period and frequency. This relationship is a core part of wave dynamics that helps us calculate one if we know the other.

frequency and period relationship

Frequency (\( f \)) and period (\( T \)) have an inverse relationship. This means if you know one, you can easily find the other. The frequency of a wave is the number of cycles per second, measured in Hertz (Hz). For a wave with a period of 34 ms, the frequency can be calculated as:
\( f = \frac{1}{34 \times 10^{-3} \text{ s}} = 29.41 \text{ Hz} \)
This relationship is crucial to understanding how wave behaviors are interrelated. When the frequency increases, the period decreases, and vice versa. This fundamental principle applies across all wave types, including sound and electromagnetic waves.

perceived frequency change

The perceived frequency change happens due to the Doppler effect, which occurs when either the source of the wave or the observer is moving. In our example, the Doppler shift doubles the perceived frequency. We calculate this new frequency (\( f' \)) as:
\( f' = 2 \times 29.41 \text{ Hz} = 58.82 \text{ Hz} \)
This perceived change affects the period of the wave as well. Using the relationship between period and frequency, the new period (\( T' \)) becomes:
\( T' = \frac{1}{f'} = \frac{1}{58.82 \text{ Hz}} = 17 \text{ ms} \)
Understanding the perceived frequency change is key in contexts like radar speed detection, astronomical observations, and medical imaging.

Doppler shift effects

The Doppler effect describes the change in frequency or wavelength of a wave relative to an observer moving relative to the wave source. This effect explains why an ambulance siren sounds higher-pitched as it approaches you and lower as it moves away. In our case, the Doppler shift effect causes the frequency of the wave to double, which in turn halves the period to 17 ms.
The Doppler effect can be seen in:

• Sound waves, as with sirens or moving vehicles
• Light waves, seen in astronomy with redshift and blueshift
• Radar, used for speed detection in law enforcement

The choices in the exercise also reflect how different relative movements (like the source or detector moving) contribute to this effect. This fundamental principle helps us understand wave behavior in various real-world scenarios.