Question

Consider a sound wave (that is, a longitudinal displacement wave) in an elastic medium with Young's modulus \(Y\) (solid) or bulk modulus \(B\) (fluid) and unperturbed density \(\rho_{0}\). Suppose this wave is described by the wave function \(\delta x(x, t),\) where \(\delta x\) denotes the displacement of a point in the medium from its equilibrium position, \(x\) is position along the path of the wave at equilibrium, and \(t\) is time. The wave can also be regarded as a pressure wave, described by wave function \(\delta p(x, t),\) where \(\delta p\) denotes the change of pressure in the medium from its equilibrium value. a) Find the relationship between \(\delta p(x, t)\) and \(\delta x(x, t),\) in general. b) If the displacement wave is a pure sinusoidal function, with amplitude \(A\), wave number \(\kappa,\) and angular frequency \(\omega,\) given by \(\delta x(x, t)=A \cos (\kappa x-\omega t)\) what is the corresponding pressure wave function, \(\delta p(x, t) ?\) What is the amplitude of the pressure wave?

Short answer

(a) The relationship between the change in pressure \(\delta p(x, t)\) and the displacement \(\delta x(x, t)\) in the elastic medium can be described by the equation \(\delta p(x, t) = -M \frac{\delta x(x, t)}{x}\), where M represents either Young's modulus (Y) for solids or the bulk modulus (B) for fluids. (b) For the given sinusoidal displacement wave function \(\delta x(x, t) = A\cos(\kappa x - \omega t)\), the corresponding pressure wave function is \(\delta p(x, t) = -M \frac{A\cos(\kappa x - \omega t)}{x}\). The amplitude of the pressure wave is given by \(-M \frac{A}{x}\).

Step-by-step solution

Relationship between stress and strain in elastic mediums

In general, for an elastic medium, the stress is related to the strain by either Young's modulus (Y) for solids or the bulk modulus (B) for fluids. In other words, stress = modulus × strain. The stress in this case is given by the change in pressure, \(\delta p\), and the strain is represented by the fractional length change, \(\frac{\delta x}{x}\). Therefore: $$\delta p = -Y \frac{\delta x}{x} \text{ (for solids)}$$ $$\delta p = -B \frac{\delta x}{x} \text{ (for fluids)}$$ Since both cases have a similar relationship, we can represent them using a single equation, replacing Y or B with M (modulus), so we have: $$\delta p = -M \frac{\delta x}{x}$$

Finding the relationship between \(\delta p(x, t)\) and \(\delta x(x, t)\)

Now we have established the relationship between \(\delta p\) and \(\delta x\) as \(\delta p = -M \frac{\delta x}{x}\), we can simply substitute the wave functions \(\delta p(x, t)\) and \(\delta x(x, t)\) to find their relationship. $$\delta p(x, t) = -M \frac{\delta x(x, t)}{x}$$

Substituting the given sinusoidal function for the displacement wave

For part (b), we are given the displacement wave function as \(\delta x(x, t) = A\cos(\kappa x - \omega t)\). We can substitute this function into the relationship we've derived in step 2 to find the corresponding pressure wave function \(\delta p(x, t)\): $$\delta p(x, t) = -M \frac{A\cos(\kappa x - \omega t)}{x}$$

Finding the amplitude of the pressure wave

The amplitude of the pressure wave is equal to the maximum value of the \(\delta p(x, t)\) function. For the given function, the maximum value of \(\delta p(x, t)\) occurs when the cosine term is at its maximum value of 1. Thus, the amplitude of the pressure wave is given by: $$\text{Amplitude of pressure wave} = -M \frac{A}{x}$$