The simultaneous application of signals $x (t)$ and $y (t)$ to the horizontal and vertical plates, respectively, of an oscilloscope, produces a vertical figure-of-8 display. If P and Q are constants and $x (t) = P s i n (4 t + 30),$ then $y (t)$ is equal to

Q s i n (4 t - 30)

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Q c o s (2 t + 30)

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Q s i n (8 t + 60)

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Q s i n (4 t + 30)

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Solution

The correct option is B $Q c o s (2 t + 30)$
$f_{y} = F r e q u e n c y o f s i g n a l a p p l i e d t o y p l a t e s .$

$f_{x} = F r e q u e n c y o f s i g n a l a p p l i e d t o X p l a t e s .$

$x (t) = P s i n (4 t + 30)$

So,
$f_{x} = \frac{4}{2 π} H z$

$\frac{f_{y}}{f_{x}} = \frac{N o . o f t i m e t a n g e n t t o u c h e s t o p o r b o t t o m}{N o . o f t i m e s t a n g e n t t o u c h e s e i t h e r s i d e} = \frac{1}{2}$

$f_{y} = \frac{f_{x}}{2} = (\frac{4}{2 π}) \times \frac{1}{2} = \frac{2}{2 π}$

$ω_{y} = 2 π f_{y} = 2 π (\frac{2}{2 π}) = 2 r a d / s e c .$

$Q s i n (2 t + 30) h a s a n g u l a r f r e q u e n c y$

$ω = 2 r a d / s e c$

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