Verified Solution • 2025–26 Session moderate

QJEEM11Ph9fff: JEE Main 2011 Physics - Class 11 JEEM11Ph9fff Solution: The transverse displacement of a wave on a string is given by…

The transverse displacement y(x,t)y(x, t) of a wave on a string is given by y(x,t)=e−(ax2+bt2+2ab xt).y\left( {x,t} \right) = {e^{ - \left( {a{x^2} + b{t^2} + 2\sqrt {ab} \,xt} \right)}}. This represents a:a:

Multiple Choice Options

  • A wave moving in −x-x direction with speed ba\sqrt {{b \over a}}
  • B standing wave of frequency b\sqrt b
  • C standing wave of frequency 1b{1 \over {\sqrt b }}
  • D wave moving in +x+x direction speed ab\sqrt {{a \over b}}

Step-by-Step Verified Solution

Given wave equation is

y(x,t)=e(−ax2+bt2+2ab xt)y\left( {x,t} \right){ = _e}\left( { - a{x^2} + b{t^2} + 2\sqrt {ab} \,xt} \right)

=e−[(ax)2+(bt)2+2ax.bt]= {e^{ - \left[ {{{\left( {\sqrt {ax} } \right)}^2} + {{\left( {\sqrt {bt} } \right)}^2} + 2\sqrt a x.\sqrt b t} \right]}}

=e−(ax+bt)2= {e^{ - {{\left( {\sqrt a x + \sqrt b t} \right)}^2}}}

=e−(x+bat)2= {e^{ - {{\left( {x + \sqrt {{b \over a}} t} \right)}^2}}}

It is a function of type y=f(x+vt)y = f\left( {x + vt} \right)

⇒\Rightarrow Speed of wave =ba= \sqrt {{b \over a}}

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