The displacements of two particles executing SHM on the same line are given as y1-asin -2t- - and y2-bsin 2-3t- - The phase difference between them at t-1 s is-

The correct option is D π6 At t=1 sec, y_1=asin(π /2+ϕ) .....(1) y_2=bsin(2π/3+ϕ)=bsin(π2+π6+ϕ)... (2)From (1) and (2) it is clear that, #160;Ph ...

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Wave Equation

The displacem...

Question

The displacements of two particles executing $S H M$ on the same line are given as $y_{1} = a sin (\frac{π}{2} t + ϕ)$ and $y_{2} = b sin (\frac{2 π}{3} t + ϕ)$ . The phase difference between them at $t = 1 s$ is:

π

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\frac{π}{2}

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\frac{π}{4}

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\frac{π}{6}

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y_{1} = a sin [(\frac{π}{2}) t + ϕ]

y_{2} = b sin [(\frac{2 π}{3}) t + ϕ]

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}

y_{1} = 0.1 s i n (100 π t + \frac{π}{3})

y_{2} = 0.1 c o s 100 π t

y_{1} = 4 sin (10 t + ϕ)

y_{2} = 5 cos (10 t)

The phase difference between two SHM

$y_{1} = 10 s i n (10 π t + \frac{π}{3}) a n d y_{2} = 12 s i n (8 π t + \frac{π}{4}) a t t = 0.5 s i s :$

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