Explanation
It is given that,
$$y=0.4\sin \left( \dfrac{440}{7}t+0.61x \right)$$
From the above equation:
$$ \omega =\dfrac{440}{7} $$
$$ \dfrac{2\pi }{T}=\dfrac{440}{7} $$
$$ \dfrac{2\times 22}{7T}=\dfrac{440}{7} $$
$$ T=0.1\,s $$
Hint:- Two wavelength passes through a point in given time so apply formula of velocity of wave accordingly.
Step 1: Note the Given values
Two consecutive crest =$$5\,m$$.
Two wave cross in time, $$t=\,1\sec $$
Wavelength = distance between two consecutive crest
$$\lambda =5\,m$$
Step 2: Calculate velocity of wave
\Velocity, $$v=\dfrac{2\lambda }{t}=\dfrac{2\times 5}{1}=10\,m{{s}^{-1}}$$
Hence, wave speed is $$10\,m{{s}^{-1}}$$
Hint:
If two waves of intensity $${{I}_{1}}\,and\,{{I}_{2}}$$ interfere, then the ration of maximum to minimum intensity in interference is given as,
$$\dfrac{{{I}_{\max }}}{{{I}_{\min }}}={{\left( \dfrac{\sqrt{{{I}_{1}}}+\sqrt{{{I}_{2}}}}{\sqrt{{{I}_{1}}}-\sqrt{{{I}_{2}}}} \right)}^{2}}$$
Step 1: Assume values of intensities.
Given,
Ratio of Intensities of two waves $${{I}_{1}}\,and\,{{I}_{2}}$$
$${{I}_{1}}:{{I}_{2}}=9:1$$.
$$\Rightarrow {{I}_{1}}=9x,\,{{I}_{2}}=1x$$
for any constant $$x$$.
Step 2: Calculate ratio of intensities.
The ratio of maximum to minimum intensity in interference is given as,
$$\Rightarrow \dfrac{{{I}_{\max }}}{{{I}_{\min }}}={{\left( \dfrac{\sqrt{9x}+\sqrt{1x}}{\sqrt{9x}-\sqrt{1x}} \right)}^{2}}$$
$$\Rightarrow \dfrac{{{I}_{\max }}}{{{I}_{\min }}}={{\left( \dfrac{4}{2} \right)}^{2}}$$
$$\Rightarrow \dfrac{{{I}_{\max }}}{{{I}_{\min }}}=\dfrac{4}{1}$$
The required ratio is 4:1. Option C.
Points B and F are in same phase as they are $$\lambda $$ distance apart.
$$\lambda $$ is the Wavelength, distance between corresponding points of two consecutive waves.
Corresponding points are those points that have completed identical fractions of their periodic motion.
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