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Chapter 7: Problem 31

Two waves, travelling in the same direction through the same region, have equal frequencies, wavelengths and amplitudes. If the amplitude of each wave is \(4 \mathrm{~mm}\) and the phase difference between the waves is \(90^{\circ}\), what is the resultant amplitude?

Short Answer

Expert verified
The resultant amplitude is approximately 5.7 mm.

Step by step solution

01

Understanding Superposition of Waves

When two waves of the same frequency and amplitude travel in the same direction, the resultant wave is formed by the superposition of the two waves. This means the amplitude of the resultant wave depends on the amplitude of the individual waves and the phase difference between them.
02

Calculate the Resultant Amplitude

The formula to determine the resultant amplitude of two waves is given by: \( A_{ ext{resultant}} = 2A\cos\left(\frac{\Delta \phi}{2}\right) \), where \( A \) is the amplitude of each wave and \( \Delta \phi \) is the phase difference. Here, \( A = 4 \mathrm{~mm} \) and \( \Delta \phi = 90^\circ \). First, convert \( \Delta \phi \) to radians for calculation: \( \Delta \phi = 90^\circ = \frac{\pi}{2} \) radians.
03

Apply the Formula

Substitute the values into the formula: \[ A_{ ext{resultant}} = 2 \times 4 \mathrm{~mm} \times \cos\left(\frac{\pi}{4}\right) \]. Calculate \( \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \).
04

Simplify the Expression

Now compute: \[ A_{ ext{resultant}} = 8 \mathrm{~mm} \times \frac{\sqrt{2}}{2} = 8 \times \frac{1.414}{2} \].
05

Final Calculation

The final calculation yields: \( A_{ ext{resultant}} = 8 \times 0.707 = 5.656 \mathrm{~mm} \). Thus, the resultant amplitude of the two waves is approximately \( 5.7 \mathrm{~mm} \).

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Key Concepts

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

Resultant Amplitude
When discussing wave superposition, the concept of resultant amplitude is crucial. It refers to the amplitude of the wave that results when two or more waves overlap. In our specific scenario, two waves with the same frequency and amplitude travel in the same direction and interfere. The resultant amplitude is influenced by the individual amplitudes and the phase difference between the waves.

The mathematical expression for resultant amplitude is given by the formula:
  • \( A_{\text{resultant}} = 2A\cos\left(\frac{\Delta \phi}{2}\right) \)
Where:
  • \( A \) is the amplitude of each wave
  • \( \Delta \phi \) is the phase difference between the waves
This equation highlights that the resultant amplitude is dependent not only on the amplitudes of the individual waves but also the phase difference between them.
Phase Difference
Phase difference is a term used to describe the difference in phase angle between two waveforms. In wave superposition, understanding phase difference is essential as it determines how waves interact with each other.

In simpler terms, the phase difference tells us how "out of step" two waves are. For instance, in the problem we are examining, the phase difference is given as \(90^{\circ}\).
  • This means one wave is a quarter of a cycle behind or ahead of the other.
  • Converting this to radians, we have \( \Delta \phi = \frac{\pi}{2} \text{ radians} \).
The phase difference affects the resultant amplitude, as it determines how much the overlapping waves reinforce or cancel each other.
Cosine Function
The cosine function, abbreviated as \( \cos \), is a fundamental mathematical function that's used extensively in wave calculations. When evaluating the resultant amplitude in wave superposition, this function becomes very important.

The formula \( A_{\text{resultant}} = 2A\cos\left(\frac{\Delta \phi}{2}\right) \) includes the cosine of half the phase difference. Here鈥檚 why:
  • The cosine function helps quantify how much two waves reinforce each other based on their phase difference.
  • For \( \Delta \phi = 90^{\circ} \), computing \( \cos \left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \) shows how much the wave amplitudes combine.
  • The value of \( \cos \) can range from -1 to 1, influencing how strongly waves add together or cancel each other out.
In this instance, a cosine value of \( \frac{\sqrt{2}}{2} \) indicates constructive interference, resulting in a partial enhancement of the resultant wave amplitude.

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Most popular questions from this chapter

A transverse wave described by $$ y=(0 \cdot 02 \mathrm{~m}) \sin \left[\left(1 \cdot 0 \mathrm{~m}^{-1}\right) x+\left(30 \mathrm{~s}^{-1}\right) t\right] $$ propagates on a stretched string having a linear mass density of \(1 \cdot 2 \times 10^{-4} \mathrm{~kg} \mathrm{~m}^{-1}\). Find the tension in the string.

A circular road of radius \(50 \mathrm{~m}\) has the angle of banking equal to \(30^{\circ}\). At what speed should a vehicle go on this road so that the friction is not used?

A wave pulse is travelling on a string with a speed \(v\) towards the positive \(X\) -axis. The shape of the string at \(t=0\) is given by \(g(x)=A \sin (x / a)\), where \(A\) and \(a\) are constants. (a) What are the dimensions of \(A\) and \(a ?\) (b) Write the equation of the wave for a general time \(t\), if the wave speed is \(v\)

Figure (15-E10) shows an aluminium wire of length \(60 \mathrm{~cm}\) joined to a steel wire of length \(80 \mathrm{~cm}\) and stretched between two fixed supports. The tension produced is \(40 \mathrm{~N}\). The cross-sectional area of the steel wire is \(1 \cdot 0 \mathrm{~mm}^{2}\) and that of the aluminium wire is \(3 \cdot 0 \mathrm{~mm}^{2}\). What could be the minimum frequency of a tuning fork which can produce standing waves in the system with the joint as a node? The density of aluminium is \(2 \cdot 6 \mathrm{~g} \mathrm{~cm}^{-3}\) and that of steel is \(7 \cdot 8 \mathrm{~g} \mathrm{~cm}^{-3}\).

A heavy ball is suspended from the ceiling of a motor car through a light string. A transverse pulse travels at a speed of \(60 \mathrm{~cm} \mathrm{~s}^{-1}\) on the string when the car is at rest and \(62 \mathrm{~cm} \mathrm{~s}^{-1}\) when the car accelerates on a horizontal road. Find the acceleration of the car. Take \(g=10 \mathrm{~m} \mathrm{~s}^{-2}\).
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