Problem 34 - College Physics

Question

A teacher standing in a doorway \(1.0 \mathrm{~m}\) wide blows a whistle with a frequency of \(1000 \mathrm{~Hz}\) to summon children from the playground (v Fig. 24.31 ). Two boys are playing on the swings \(20 \mathrm{~m}\) away from the school building. One boy is at an angle of \(0^{\circ}\) and another one at \(19.6^{\circ}\) from a line normal to the doorway. Taking the speed of sound in air to be \(335 \mathrm{~m} / \mathrm{s}\), which boy may not hear the whisle? Prove your answer.

Step-by-Step Solution

Answer

The second boy at an angle of \(19.6^{\circ}\) may not hear the whistle due to diffraction.

Step 1: Identify Information and Write down Given Data

We are given:- Frequency of the whistle, \(f = 1000 \text{ Hz}\).- Speed of sound, \(v = 335 \text{ m/s}\).- Width of doorway, \(a = 1.0 \text{ m}\).- Distance of the boys from the doorway, \(D = 20 \text{ m}\).- Angles of the boys from the normal to the doorway: \(\theta_1 = 0^{\circ}\) and \(\theta_2 = 19.6^{\circ}\).

Step 2: Calculate Wavelength of Sound

The wavelength \(\lambda\) of the sound can be calculated from the frequency \(f\) and speed of sound \(v\) using the formula:\[ \lambda = \frac{v}{f} \]Substitute the values:\[ \lambda = \frac{335}{1000} = 0.335 \text{ m} \]

Step 3: Determine the First Minimum Angle Using Single Slit Diffraction

For single slit diffraction, the first minimum occurs at angle \(\theta\) given by:\[ a \sin(\theta) = m \lambda \]where \(m = 1\) for the first minimum. Substitute \(a = 1.0 \text{ m}\) and \(\lambda = 0.335 \text{ m}\):\[ (1.0) \sin(\theta) = 1 \times 0.335 \]\[ \sin(\theta) = 0.335 \]\[ \theta = \sin^{-1}(0.335) \approx 19.58^{\circ} \]

Step 4: Compare Boys' Angles to Diffraction Minimum Angle

The angle for the first minimum results in sound being very weak and potentially inaudible. Compare the given angles of the boys with the calculated diffraction angle:- Boy 1: \(\theta_1 = 0^{\circ}\)- Boy 2: \(\theta_2 = 19.6^{\circ}\)Boy 2's angle \(19.6^{\circ}\) is slightly greater than \(19.58^{\circ}\), so diffraction might cause the whistle's sound to be barely audible or not heard by Boy 2. Boy 1 is at an angle where the sound is not weak by diffraction.

Key Concepts

Sound Wave Diffraction

Sound wave diffraction is a fascinating phenomenon that occurs when sound waves encounter an obstacle or a slit that is comparable in size to their wavelength. This process can cause the sound waves to bend around the obstacles, spreading into the region beyond them.
In the context of the exercise, when a teacher blows a whistle through a doorway, the sound waves pass through this opening which acts as a single slit. The width of the doorway becomes crucial because if the slit width and the wavelength of the sound wave are similar, significant diffraction occurs.

• This diffraction generates patterns of maxima and minima in the intensity of sound.
• The first minimum occurs when sound waves are most significantly bent, causing a zone of reduced sound intensity.

When a sound wave diffracts, the energy is spread across various directions, which can make the sound much weaker in some directions. This is precisely what affects whether the boys can hear the whistle from different angles.

Wave Interference

Wave interference is another key concept here. As sound waves diffract through the doorway, they can interfere with each other.
Interference can be either constructive or destructive.

• Constructive interference amplifies sounds when incoming waves combine to produce a wave of greater amplitude.
• Destructive interference happens when waves combine to lower their overall amplitude.

Destructive interference is what mostly contributes to sound fading or becoming undetectable. In the case of Boy 2, positioned at an angle where interference causes minima due to diffraction, he may find the sound of the whistle too weak.
By understanding how sound waves interact through diffraction and interference, one can determine where and when sound might be heard or missed.

Physics Problem Solving

Physics problem solving requires a strategic approach. Here's how to tackle problems involving sound wave diffraction:

• **Identify given data:** Start by writing down all the known values, such as frequency, speed of sound, and dimensions of the slit.
• **Apply relevant formulas:** Use physics formulas, like the one for the diffraction angle \(a \sin(\theta) = m \lambda\), to find unknown quantities.
• **Calculate step-by-step:** Break down complex problems into smaller tasks to avoid errors. For instance, calculate the wavelength before determining diffraction angles.
• **Analyze results:** Compare your findings with the problem statement to draw conclusions, just as we identified which boy might not hear the sound based on the angle.

This method not only helps in solving physics problems related to sound wave diffraction but also enhances comprehension of physics concepts and their applications in real-world scenarios.