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

A plane wave travelling along the X-direction with unit amplitude and wavelength \(\lambda\) can be written, at \(t=0\), as a displacement field

\[u(x,0) = A\sin\left(\frac{2\pi x}{\lambda}\right), \qquad A=1\]

Step 1: In a 1-D wave, strain is the spatial gradient of displacement:

\[\varepsilon(x) = \frac{\partial u}{\partial x} = A\frac{2\pi}{\lambda}\cos\left(\frac{2\pi x}{\lambda}\right)\]

Step 2: The strain is maximum where the cosine term equals 1 (at the zero-crossings of the displacement, where the wave's local slope is steepest), giving

\[\varepsilon_{max} = \frac{2\pi A}{\lambda}\]

Step 3: Substituting \(A=1\) and \(\lambda=16\) km,

\[\varepsilon_{max} = \frac{2\pi(1)}{16} = \frac{\pi}{8} \approx 0.3927\]

Rounded to three decimal places, \(\varepsilon_{max}\approx 0.393\), which lies within the accepted range (0.385 to 0.405).

\(\boxed{\varepsilon_{max}\approx 0.393}\)