Question

The maximum strain for the plane wave at \(t=0\), having a wavelength of 16 km and unit amplitude, travelling along the X-direction, as shown in the figure, is _______ (rounded off to three decimal places).

Hint:

Strain is the spatial derivative of displacement; for a sinusoid the peak strain equals amplitude times wavenumber (2 pi / lambda).

Answer 1

Correct Answer: 0.385

Define the wavenumber \(k=2\pi/\lambda\), so the wave profile at \(t=0\) is \(u(x)=A\sin(kx)\).

Strain is the fractional deformation, \(\varepsilon = du/dx = Ak\cos(kx)\), whose peak value is simply \(Ak\) since \(\cos(kx)\) oscillates between \(-1\) and \(+1\).

With unit amplitude \(A=1\) and \(\lambda=16\) km,

\[k = \frac{2\pi}{16} = \frac{\pi}{8}\ \text{km}^{-1} \approx 0.3927\ \text{km}^{-1}\]

Because the amplitude is dimensionless (unit amplitude), the peak strain magnitude here equals \(k\) numerically, i.e. \(\varepsilon_{max}=\pi/8\approx 0.393\).

This can also be checked qualitatively: doubling the wavelength while keeping the same peak-to-trough displacement spreads that same displacement over twice the distance, halving the maximum strain - consistent with \(\varepsilon_{max}\propto 1/\lambda\).

\(\boxed{\varepsilon_{max}\approx 0.393}\), within the accepted band 0.385 to 0.405.