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Q.

A simple harmonic wave has the equation $y_{1} = 0.3 \sin (314 t - 1.57 x)$ and another wave has equation $y_{2} = 0.1 \sin (314 t - 1.57 x + 1.57)$ where $x, y_{1}$ and $y_{2}$ are in metre and t is in second.

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a

$λ_{1} = λ_{2} = 4 m$

b

$v_{1} = v_{2} = 50 H z$

c

Ratio of intensity is 9

d

$y_{2}$ leads $y_{1}$ by a phase angle of $\frac{π}{2}$

answer is A, B, C, D.

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Detailed Solution

$ω_{1} = ω_{2} = 314 r a d / \sec$

$\begin{array}{l} \Rightarrow 2 π v_{1} = 2 π v_{2} = 314 \\ \Rightarrow v_{1} = v_{2} = 50 H z and \\ K_{1} = K_{2} = 1.57 = \frac{π}{2} \Rightarrow \frac{2 π}{λ_{1}} = \frac{2 π}{λ_{2}} = \frac{π}{2} \\ \Rightarrow λ_{1} = λ_{2} = 4 m . \frac{I_{1}}{I_{2}} = \frac{a_{1}^{2}}{a_{2}^{2}} = \frac{0.09}{0.01} = 9 \\ Y_{2} {leads Y}_{1} by 1.57 radian= \frac{π}{2} \end{array}$

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