JEE MAIN - Physics (2008 - No. 18)
A wave travelling along the $$x$$-axis is described by the equation $$y(x, t)=0.005$$ $$\cos \,\left( {\alpha \,x - \beta t} \right).$$ If the wavelength and the time period of the wave are $$0.08$$ $$m$$ and $$2.0s$$, respectively, then $$\alpha $$ and $$\beta $$ in appropriate units are
$$\alpha = 25.00\pi ,\,\beta = \pi $$
$$\alpha = {{0.08} \over \pi },\,\beta = {{2.0} \over \pi }$$
$$\alpha = {{0.04} \over \pi },\,\beta = {{1.0} \over \pi }$$
$$\alpha = 12.50\pi ,\,\beta = {\pi \over {2.0}}$$
Explanation
$$y\left( {x,t} \right) = 0.005\,\cos \left( {\alpha x - \beta t} \right)$$ (Given)
Comparing it with the standard equation of wave
$$y\left( {x,t} \right) = a\cos \left( {kx - \omega t} \right)$$ we get
$$k = \alpha $$ $$\,\,\,\,\,$$ and $$\,\,\,\,\,$$ $$\omega = \beta $$
$$\therefore$$ $${{2\pi } \over \gamma } = \alpha $$ $$\,\,\,\,\,$$ and $$\,\,\,\,\,$$ $${{2\pi } \over T} = \beta $$
$$\therefore$$ $$\alpha = {{2\pi } \over {0.08}} = 25\pi $$ $$\,\,\,\,\,$$ and $$\,\,\,\,\,$$ $$\beta = {{2\pi } \over 2} = \pi $$
Comparing it with the standard equation of wave
$$y\left( {x,t} \right) = a\cos \left( {kx - \omega t} \right)$$ we get
$$k = \alpha $$ $$\,\,\,\,\,$$ and $$\,\,\,\,\,$$ $$\omega = \beta $$
$$\therefore$$ $${{2\pi } \over \gamma } = \alpha $$ $$\,\,\,\,\,$$ and $$\,\,\,\,\,$$ $${{2\pi } \over T} = \beta $$
$$\therefore$$ $$\alpha = {{2\pi } \over {0.08}} = 25\pi $$ $$\,\,\,\,\,$$ and $$\,\,\,\,\,$$ $$\beta = {{2\pi } \over 2} = \pi $$
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