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 +-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?

Verified step by step guidance

1. First, determine the frequency of the waves. Since 75 wave crests pass by each minute, convert this to seconds to find the frequency: \( f = \frac{75}{60} \text{ Hz} \).
2. Next, calculate the wavelength of the waves using the wave speed formula: \( v = f \lambda \), where \( v = 15.0 \text{ cm/s} \). Rearrange to find the wavelength: \( \lambda = \frac{v}{f} \).
3. Understand that the problem involves diffraction, where waves spread out after passing through a hole. The angle \( \theta \) to the first minimum in the diffraction pattern is given by \( \sin \theta = \frac{\lambda}{a} \), where \( a \) is the width of the hole.
4. The distance from the point directly opposite the hole to the first minimum on the shore is given as \( \pm 61.3 \text{ cm} \). Use the small angle approximation \( \tan \theta \approx \sin \theta \approx \theta \) to relate this distance to the angle: \( \theta = \frac{61.3 \text{ cm}}{320 \text{ cm}} \).
5. Finally, solve for the width of the hole \( a \) using the equation \( \sin \theta = \frac{\lambda}{a} \). Substitute the values for \( \lambda \) and \( \theta \) to find \( a \).

Key Concepts

Wave Speed

Wave speed is the rate at which a wave propagates through a medium, calculated as the product of wavelength and frequency. In this problem, the wave speed is given as 15.0 cm/s, which helps determine the wavelength when combined with the frequency of wave crests passing a point.

Diffraction

Diffraction refers to the bending and spreading of waves when they encounter an obstacle or pass through a narrow opening. The extent of diffraction depends on the size of the opening relative to the wavelength. In this scenario, diffraction explains why waves reach the shore within a certain distance from the hole.

Wave Frequency

Wave frequency is the number of wave crests passing a point per unit time, measured in hertz (Hz). Here, 75 wave crests pass by each minute, translating to a frequency of 1.25 Hz. This frequency, along with the wave speed, is crucial for calculating the wavelength, which is needed to determine the width of the hole.