Problem 7 - University Physics with Modern Physics

Question

A series of parallel linear water wave fronts are traveling directly toward the shore at 15.0 cm/s on an otherwise placid lake. A long concrete barrier that runs parallel to the shore at a distance of 3.20 m away has a hole in it. You count the wave crests and observe that 75.0 of them pass by each minute, and you also observe that no waves reach the shore at \(\pm\)61.3 cm from the point directly opposite the hole, but waves do reach the shore everywhere within this distance. (a) How wide is the hole in the barrier? (b) At what other angles do you find no waves hitting the shore?

Step-by-Step Solution

Answer

Hole width is 67.35 cm; no waves at 20.9°, 30.6°, etc.

Step 1: Calculate Wave Frequency and Wavelength

We know the velocity \( v = 15.0 \) cm/s and the number of crests passing per minute is 75.0. First, convert 75.0 crests/min to crests/s: \( \frac{75.0}{60} = 1.25 \) crests/s. This is the frequency \( f \). The wavelength \( \lambda \) can be found using the formula \( v = f \lambda \). Thus, \( \lambda = \frac{v}{f} = \frac{15.0}{1.25} = 12.0 \) cm.

Step 2: Use Diffraction Angle Formula

The given problem involves diffraction. The condition for the first diffraction minimum (where no waves reach) is given by \( a \sin \theta = m \lambda \), where \( a \) is the width of the opening, \( m \) is an integer order (\( m = \pm 1, \pm 2, \ldots \)), and \( \theta \) is the angle from the central line (90 degrees or directly opposite the hole). Here, \( m = 1 \), and \( \theta = \pm \tan^{-1}(\frac{61.3}{320}) \) to approximate the angle to the shore. This yields \( \theta \approx 10.9^\circ \).

Step 3: Solve for the Width of the Hole

Substituting the known values into the diffraction formula, with \( a \sin \theta = \pm 1 \times 12.0\, \text{cm} \), where \( \sin \theta = \frac{61.3}{320} \), solve: \( a = \frac{12.0}{\sin(10.9)} \approx 67.35 \) cm.

Step 4: Identify Other Angles for No Waves

For higher orders \( m = 2, 3, \ldots \), use \( a \sin \theta = m \lambda \). Since \( a = 67.35 \) cm and \( \lambda = 12 \) cm, solve for \( \theta \). For \( m = 2\), \( \sin \theta = \frac{2 \times 12.0}{67.35} \approx 0.356 \), thus \( \theta \approx 20.9^\circ \). Continue similarly for higher integers.

Key Concepts

Wave Frequency

Wave frequency refers to the number of wave crests passing a given point per unit time. In the context of water waves traveling towards a shore, knowing the frequency helps us understand how often we can expect wave crests to arrive. In the given exercise, we observe 75 wave crests passing every minute. To find the frequency in a more standard scientific unit, such as Hertz (Hz), which is the same as waves per second, we convert this number by dividing by 60. This calculation gives us:

• Frequency (\( f \)) = 1.25 waves per second (1.25 Hz).

Knowing the frequency is crucial for understanding and analyzing the behavior of waves as they encounter obstacles like the concrete barrier in the exercise. This parameter plays a key role in calculating other properties of the wave, such as its wavelength, using the velocity of the wave fronts.

Wavelength Calculation

Wavelength is the distance between successive crests of a wave. To find the wavelength of the water waves in this exercise, we use the relationship between the wave's speed (\( v \)), its frequency (\( f \)), and its wavelength (\( \lambda \)) which can be given by the formula:

• \( v = f \lambda \)
• \( \lambda = \frac{v}{f} \)

Substitute the known values, where the velocity is 15 cm/s and the frequency is 1.25 Hz:

• \( \lambda = \frac{15.0}{1.25} \approx 12.0 \text{ cm} \)

Understanding how to compute the wavelength is essential as it influences how the wave will diffract or bend around obstacles, affecting which areas it can reach beyond the barrier.

Diffraction Formula

Diffraction occurs when a wave encounters an obstacle or slit's edge, bending around it. The extent to which waves diffract depends on their wavelength and the size of the opening through which they pass. The formula for the first diffraction minimum is given by:

• \( a \sin \theta = m \lambda \)

where:

• \( a \)is the width of the hole,
• \( \theta \)is the angle relative to the normal direction (how far the wave diffracts),
• \( m \)is the order of the minima (\( m = \pm 1, \pm 2, \ldots \)), and
• \( \lambda \)is the wavelength.

In the exercise, to find the hole's width where no waves reach, we set \( m = 1 \) for the first minimum, meaning:

• \( a = \frac{12.0}{\sin(\theta)} \)

This tells us the critical angle at which waves stop hitting the shore, helping analyze further angles causing similar behavior.

Diffraction Angles

Diffraction angles are the angles at which wave intensity reaches zero after passing through a slit or past an edge. This is crucial for determining where waves will not reach a surface beyond an obstacle. By using the formula:

• \( a \sin \theta = m \lambda \)

we apply varying integer values of \( m \) to find multiple angles (\( \theta \)), resulting in points of minimum intensity on the opposite side.For the first minimum with \( m = 1 \), \( \theta \approx 10.9^\circ \), using \( \sin \theta = \frac{61.3}{320} \).Other minima are found by increasing \( m \):

• For \( m = 2 \), \( \sin \theta = \frac{2 \times 12}{67.35} \approx 0.356 \), or \( \theta \approx 20.9^\circ \).
• Continue similarly with \( m > 2 \) for additional angles.

This understanding allows for predicting where to expect calm seas on the shore behind such barriers.