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Chapter 7: Problem 33

Two waves, each having a frequency of \(100 \mathrm{~Hz}\) and a wavelength of \(2 \cdot 0 \mathrm{~cm}\), are travelling in the same direction on a string. What is the phase difference between the waves (a) if the second wave was produced \(0-015\) s later than the first one at the same place, \((b)\) if the two waves were produced at the same instant but the first one was produced a distance \(4 \cdot 0 \mathrm{~cm}\) behind the second one? (c) If each of the waves has an amplitude of \(2 \cdot 0 \mathrm{~mm}\), what would be the amplitudes of the resultant waves in part (a) and (b)?

Short Answer

Expert verified
(a) Phase difference: \(3\pi\), amplitude: 0 mm; (b) Phase difference: \(4\pi\), amplitude: 4.0 mm.

Step by step solution

01

Understand Wave Properties and Definitions

We are given two waves with the same frequency and wavelength traveling in the same direction. The frequency of each wave is \( f = 100 \, \text{Hz} \) and the wavelength is \( \lambda = 2.0 \, \text{cm} = 0.02 \, \text{m} \). The speed of the wave \( v \) can be calculated by \( v = f \times \lambda = 100 \, \text{Hz} \times 0.02 \, \text{m} = 2 \, \text{m/s} \).
02

Calculate Phase Difference for Time Delay

(a) If the second wave is produced \(0.015\) seconds later, the phase difference \( \Delta \phi \) due to time delay can be calculated using \( \Delta t = 0.015 \, \text{s} \). The phase difference in terms of angle is given by \( \Delta \phi = 2\pi f \Delta t \). Substitute the given values: \( \Delta \phi = 2\pi \times 100 \, \text{Hz} \times 0.015 \, \text{s} = 3\pi \).
03

Calculate Phase Difference for Spatial Separation

(b) If the first wave is produced 4.0 cm behind, the phase difference \( \Delta \phi \) due to space can be calculated using the distance \( \Delta x = 4.0 \, \text{cm} = 0.04 \, \text{m} \). The phase difference is \( \Delta \phi = \frac{2\pi \Delta x}{\lambda} \). Substitute the values: \( \Delta \phi = \frac{2\pi \times 0.04 \, \text{m}}{0.02 \, \text{m}} = 4\pi \).
04

Calculate Resultant Amplitude for Time-Delayed Waves

The resultant amplitude for overlapping waves can be calculated using the equation: \[ A_r = 2A \cos\left(\frac{\Delta \phi}{2}\right) \]where \( A = 2.0 \, \text{mm} = 0.002 \, \text{m} \) and \( \Delta\phi = 3\pi \). Since \( \cos\left(\frac{3\pi}{2}\right) = 0 \), \( A_r = 2 \times 0.002 \, \text{m} \times 0 = 0 \, \text{m} \).
05

Calculate Resultant Amplitude for Waves with Spatial Separation

For the case of spatial separation with \( \Delta\phi = 4\pi \), the cosine term is \( \cos\left(\frac{4\pi}{2}\right) = 1 \). Using the same amplitude equation: \[ A_r = 2 \times 0.002 \, \text{m} \times 1 = 0.004 \, \text{m} = 4.0 \, \text{mm} \].
06

Conclusion

The phase difference in part (a) is \(3\pi\), and in part (b) is \(4\pi\). The amplitude of the resultant wave in part (a) is 0 mm, and in part (b) it is 4.0 mm.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Phase Difference
When two waves interfere, the phase difference between them is a crucial aspect that determines how they will combine. Phase difference refers to the shift in the cycles of two waves. It can result from either a time delay or spatial separation.
  • **Time Delay**: If one wave is generated at a different time compared to the other, a phase shift develops. You calculate this shift by multiplying the angular frequency by the time delay.
  • **Spatial Separation**: When waves start from different positions, their phase difference is dependent upon how much one wave lags behind the other in terms of the wave鈥檚 wavelength.

The calculation involves mathematical formulas that use angles represented in radians. For example, a full cycle corresponds to an angle of \(2\pi\). Understanding phase differences is essential for predicting whether waves will reinforce or cancel each other out when they meet.
Wavelength
Wavelength is a fundamental property of waves that represents the distance between successive crests (or troughs) of a wave. It is typically denoted by the Greek letter \( \lambda \). The wavelength is crucial in determining how waves interact.
  • It is inversely related to the frequency, meaning as wavelength increases, the frequency tends to decrease.
  • Wavelength is important for calculating phase differences when waves are spatially separated.

In the exercise, the wavelength is given as \(0.02\, \text{m}\) (20 cm). This value is used to calculate how much one wave lags behind another in terms of cycles or phase, depending on their starting points. In practical measurements, knowing the wavelength helps in predicting the behavior of wave interference.
Frequency
The frequency of a wave is the number of waves that pass a point in one second. It is measured in hertz (Hz) and is denoted by \(f\). A higher frequency means more waves passing by each second. In the given problem, the frequency is \(100 \mathrm{~Hz}\).
  • Frequency and wavelength are related through wave speed, which can be calculated using the formula \(v = f \times \lambda\).
  • Changes in frequency can affect the pitch of sound or color in light.

For wave interference, frequency remains constant for waves generated from the same source under the same conditions. It is critical in determining the time-based phase difference when considering how waves overlap with each other.
Resultant Amplitude
When two waves overlap, they interfere, leading to a resultant wave whose amplitude depends on their phase difference. The resultant amplitude is the measure of how "tall" the wave becomes when two waves combine.
  • Constructive interference happens when waves are in phase, leading to a higher amplitude.
  • Destructive interference occurs when waves are out of phase, possibly canceling each other out.

The formula for resultant amplitude in interfering waves is given by:\[ A_r = 2A \cos\left(\frac{\Delta \phi}{2}\right) \]where \(A\) is the amplitude of individual waves and \(\Delta \phi\) is the phase difference. In case (a) from the problem, the amplitude turns to zero because the cosine term vanishes, exemplifying perfect destructive interference. While in case (b), it reaches its maximum potential as the waves are constructively reinforced.

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Most popular questions from this chapter

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