Question #c53c7

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Question #c53c7

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Answer 1 · 2017-02-07T10:01:47

Given the common oscillator frequency is

$ν = 7.80 \times 10^{2} H z = 780 H z$ $nu=7.80xx10^2Hz=780Hz$

Velocity of sound $V = 343 m s^{- 1}$ $V=343ms^-1$

So the wave lenhth of sound created from each spreaker will be $λ = \frac{V}{ν} = \frac{343}{780} \approx 0.44 m$ $lambda=V/nu=343/780~~0.44m$

Here a stationary wave will be formed due to superposition of two similar waves approaching from opposite sides.The distance between its two consecutive nodes will be $\frac{λ}{2} = \frac{0.44}{2} m = 0.22 m$ $lambda/2=0.44/2m=0.22m$ .

If the two speakers vibrate in phase, the point halfway between them will be an antinode of pressure and the distance of this antinode point (minimum) from either of the speaker will be $= \frac{1.27}{3} m = 0.635 m$ $=1.27/3 m = 0.635m$ Let this position be named as M.

The distances of other nodes from the left side (at the left of mid position)

$L_{1} = 0.635 - \frac{λ}{4} = 0.635 - 0.11 = 0.525 m$ $L_1=0.635-lambda/4=0.635-0.11=0.525m$

$L_{2} = 0.525 - \frac{λ}{2} = 0.525 - 0.22 = 0.305 m$ $L_2=0.525-lambda/2=0.525-0.22=0.305m$

$L_{3} = 0.305 - \frac{λ}{2} = 0.305 - 0.22 = 0.085 m$ $L_3=0.305-lambda/2=0.305-0.22=0.085m$

The distances of other nodes from the left side (at the right of mid position)

$R_{1} = 0.635 + \frac{λ}{4} = 0.635 + 0.11 = 0.745 m$ $R_1=0.635+lambda/4=0.635+0.11=0.745m$

$R_{2} = 0.745 + \frac{λ}{2} = 0.745 + 0.22 = 0.965 m$ $R_2=0.745+lambda/2=0.745+0.22=0.965m$

$R_{3} = 0.965 + \frac{λ}{2} = 0.965 + 0.22 = 1.185 m$ $R_3=0.965+lambda/2=0.965+0.22=1.185m$

Hence the positions of the relative minima on the line joining the positions of two speakers will be as follows (from left)

$L_{3} = 0.085 m, L_{2} = 0.305 m, L_{2} = 0.525 m, M = 0.635 m, R_{1} = 0.745 m, R_{2} = 0.965 m, R_{3} = 1.185 m$ $L_3=0.085m,L_2=0.305m,L_2=0.525m,M=0.635m,R_1=0.745m,R_2=0.965m,R_3=1.185m$

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