If x- a sin -t-6 and y- a cos-t- then the phase difference between the two waves is

The correct option is A π 3 X= a sin (wt+π/6) Y= a sin (wt+π/2) ϕ_1=π/6 ϕ_2=π/2 ϕ=ϕ_2 ϕ_1=π/3 ...

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Wave Equation

If x= a sin...

Question

If $x = a sin (ω t + π / 6)$ and $y = a cos ω t$ , then the phase difference between the two waves is

\frac{π}{3}

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\frac{π}{6}

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\frac{π}{2}

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π

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If $x = a s i n [ω t + \frac{π}{6}] a n d x^{'} = a c o s ω t$ , then what is the phase difference between the two waves?

x_{1} = A sin (ω t + \frac{π}{6})

x_{2} = A cos ω t

If $x = a s i n (ω t + \frac{π}{6})$ and $x = acoswt$ then what is the phase difference between the two waves?

f (x) = lim n \to \infty {\frac{x}{1 + {3 {tan}^{2} x}^{n}}}

f (x)

π

f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \begin{matrix} 0, & - \frac{π}{2} < x < - \frac{π}{6} \end{matrix} \begin{matrix} \frac{x}{2}, & x = - \frac{π}{6} \end{matrix} \begin{matrix} x, & - \frac{π}{6} < x < \frac{π}{6} \end{matrix} \begin{matrix} \frac{x}{2}, & x = \frac{π}{6} \end{matrix} \begin{matrix} 0, & \frac{π}{6} < x < \frac{π}{2} \end{matrix} \end{matrix}

If $x = a s i n [ω t + \frac{π}{6}] a n d x^{'} = a c o s ω t$ , then what is the phase difference between the two waves?

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