JEE MAIN - Physics (2011 - No. 14)
The transverse displacement $$y(x, t)$$ of a wave on a string is given by $$y\left( {x,t} \right) = {e^{ - \left( {a{x^2} + b{t^2} + 2\sqrt {ab} \,xt} \right)}}.$$ This represents $$a:$$
wave moving in $$-x$$ direction with speed $$\sqrt {{b \over a}} $$
standing wave of frequency $$\sqrt b $$
standing wave of frequency $${1 \over {\sqrt b }}$$
wave moving in $$+x$$ direction speed $$\sqrt {{a \over b}} $$
Explanation
Given wave equation is
$$y\left( {x,t} \right){ = _e}\left( { - a{x^2} + b{t^2} + 2\sqrt {ab} \,xt} \right)$$
$$ = {e^{ - \left[ {{{\left( {\sqrt {ax} } \right)}^2} + {{\left( {\sqrt {bt} } \right)}^2} + 2\sqrt a x.\sqrt b t} \right]}}$$
$$ = {e^{ - {{\left( {\sqrt a x + \sqrt b t} \right)}^2}}}$$
$$ = {e^{ - {{\left( {x + \sqrt {{b \over a}} t} \right)}^2}}}$$
It is a function of type $$y = f\left( {x + vt} \right)$$
$$ \Rightarrow $$ Speed of wave $$ = \sqrt {{b \over a}} $$
$$y\left( {x,t} \right){ = _e}\left( { - a{x^2} + b{t^2} + 2\sqrt {ab} \,xt} \right)$$
$$ = {e^{ - \left[ {{{\left( {\sqrt {ax} } \right)}^2} + {{\left( {\sqrt {bt} } \right)}^2} + 2\sqrt a x.\sqrt b t} \right]}}$$
$$ = {e^{ - {{\left( {\sqrt a x + \sqrt b t} \right)}^2}}}$$
$$ = {e^{ - {{\left( {x + \sqrt {{b \over a}} t} \right)}^2}}}$$
It is a function of type $$y = f\left( {x + vt} \right)$$
$$ \Rightarrow $$ Speed of wave $$ = \sqrt {{b \over a}} $$
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