Question

The equation of a simple harmonic progressive wave is $y = 0.4\sin 100\left( {t - \dfrac{x}{{40}}} \right)$ in S.I units. Calculate wavelength.

Answer 1

A simple harmonic progressive wave is a type of wave which moves forwards continuously in a particular direction without changing its form and the particles of the medium through which the wave is travelling perform S.H.M about their mean positions with amplitude and period remaining the same.

Complete step by step solution:
Calculate the wavelength of Simple Harmonic Oscillator:
We have been given the equation of a simple harmonic progressive wave as:
$\Rightarrow y = 0.4\sin 100\left( {t - \dfrac{x}{{40}}} \right)$;
$\Rightarrow y = 0.4\sin \left( {100t - \dfrac{{100x}}{{40}}} \right)$
Now, the general equation of a wave is given by:
$\Rightarrow y = A\sin \left( {\omega t - kx} \right)$;
Now compare the last terms of both the equations:
$\Rightarrow \left( {\omega t - kx} \right) = \left( {100t - \dfrac{{100x}}{{40}}} \right)$;
By comparing the above two equations we have found out that:
$\Rightarrow \omega = 100;$
$\Rightarrow k = \dfrac{{100}}{{40}};$
The wavenumber constant “k” in relation with wavelength “$\lambda$” is given by:
$\Rightarrow k = \dfrac{{2\pi }}{\lambda };$
Equate the above two relations:
$\Rightarrow \dfrac{{2\pi }}{\lambda } = \dfrac{{100}}{{40}}$;
Solve the above relation for the wavelength “$\lambda$”.
$\Rightarrow \lambda = \dfrac{{40 \times 2\pi }}{{100}}$;
The wavelength of the Simple Harmonic progressive wave is:
$\Rightarrow \lambda = 2.51m$

Therefore, the wavelength of Simple Harmonic progressive wave $y = 0.4\sin 100\left( {t - \dfrac{x}{{40}}} \right)$is 2.51m.

Note: A simple harmonic motion is defined as a motion which is periodic in nature and where the restoring force on an object which is moving is directly proportional to the magnitude of displacement of the object. Here, we need to compare the given equation with the general equation of a wave and compare both the components of the equation together and find out the wavelength by making a relation between the wavenumber “k” and the wavelength lambda.