JEE MAIN - Physics (2018 - 15th April Evening Slot - No. 17)
Two simple harmonic motions, as shown below, are at right angles. They are combined to form Lissajous figures.
x(t) = A sin (at + $$\delta $$)
y(t) = B sin (bt)
Identify the correct match below.
x(t) = A sin (at + $$\delta $$)
y(t) = B sin (bt)
Identify the correct match below.
Parameters A $$ \ne $$ B, a = b; $$\delta $$ = 0;
Curve Parabola
Curve Parabola
Parameters A = B, a = b; $$\delta $$ = $${\raise0.5ex\hbox{$\scriptstyle \pi $}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}$$
Curve Line
Curve Line
Parameters A $$ \ne $$ B, a = b; $$\delta $$ = $${\raise0.5ex\hbox{$\scriptstyle \pi $}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}$$
Curve Ellipse
Curve Ellipse
Parameters A = B, a = 2b; $$\delta $$ = $${\raise0.5ex\hbox{$\scriptstyle \pi $}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}$$
Curve Circle
Curve Circle
Explanation
The given simple harmonic motions to form Lissajous figures are $$x(t) = A\sin (at + \delta )$$ and $$y(t) = B\sin (bt)$$.
For parabola, conditions should be
A = B or A $$\ne$$ B, a = 2b, $$\delta$$ = $$\pi$$/2
For line, conditions should be
A = B, a = b, $$\delta$$ = $$\pi$$
For circle, condition should be
A = B, a = b; $$\delta$$ = $$\pi$$/2
For ellipse, condition should be
A $$\ne$$ B, a = b; $$\delta$$ = $$\pi$$/2
Therefore, we obtain an ellipse.
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