Question

# Explain analytically how stationary waves are formed. What are nodes and antinodes? Show that the distance between two adjacent nodes or antinodes is $$\dfrac {\lambda}{2}$$

Solution

## Consider two simple harmonic progressive waves of the same amplitude $$A$$, wavelength $$\lambda$$ and frequency $$n = \omega /2\pi$$ travelling along  the x-axis in opposite directions. They may be represented by$$y_{1} = A\sin (\omega t - kx)$$ (along the $$+$$ x-axis) and $$..... (1)$$$$y_{2} = A\sin (\omega t + kx)$$ (along the $$-$$ x-axis) $$.... (2)$$where $$k = 2\pi /\lambda$$ is the propagation constant.By the superposition principle,$$y = y_{1} + y_{2} = A[\sin (\omega t - kx) + \sin (\omega t + kx)]$$Using the trigonometrical identity,$$\sin C + \sin D = 2\sin \left (\dfrac {C + D}{2}\right )\cos \left (\dfrac {C + D}{2}\right )$$$$y = 2A \sin \omega t \cos (-kx)$$$$= 2A \sin \omega t \cos kx [\cos (-kx) = \cos (kx)]$$$$= 2A \cos kx \sin \omega t .... (3)$$$$y = R\sin \omega t .... (4)$$where $$R = 2A\cos kx .... (5)$$The above equation shows that the resultant disturbance is simple harmonic having the same period as that of the individual waves and the amplitude $$R$$.The point at which the particles of the medium are always at rest are called the nodes.At nodes $$R = 0$$$$\therefore \cos \dfrac {2\pi x}{\lambda} = 0 \left (\because A\neq 0\ and \ k = \dfrac {2\pi}{\lambda}\right )$$$$\therefore \dfrac {2\pi x}{\lambda} = \dfrac {\pi}{2}, \dfrac {3\pi}{2}, \dfrac {5\pi}{2}, \dfrac {7\pi}{2}, ....$$$$\therefore x = \dfrac {\lambda}{4}, \dfrac {3\lambda}{4}, \dfrac {5\lambda}{4}, \dfrac {7\lambda}{4}, ...., (2m + 1) \dfrac {\lambda}{4} .... (6)$$where $$m = 0, 1, 2, ...$$ therefore the distance between successive nodes is$$[2(m + 1) + 1] \dfrac {\lambda}{4} - (2m + 1) \dfrac {\lambda}{4} = \dfrac {\lambda}{2}$$.The points at which the particles of the medium vibrate with the maximum amplitude are called the antinodes.At antinodes: $$R = \pm 2A$$$$\therefore \cos \dfrac {2\pi x}{\lambda} = \pm 1$$$$\therefore \dfrac {2\pi x}{\lambda} = 0, \pi, 2\pi, 3\pi, ....$$$$\therefore x = 0, \dfrac {\lambda}{2}, \lambda, \dfrac {3\lambda}{2}, ..., \dfrac {m\lambda}{2} (m = 0, 1, 2) .... (7)$$Therefore the distance between successive antinodes is$$\dfrac {(m + 1)}{2}\cdot \lambda - \dfrac {m\lambda}{2} = \dfrac {\lambda}{2}$$$$\therefore$$ Distance between successive nodes$$=$$ distance between successive antinodes $$= \dfrac {\lambda}{2}$$From eq. $$(6)$$ and $$(7)$$, it can be seen that the nodes and the antinodes occur alternately and are equally spaced.Physics

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