Two loudspeakers L_1 and L_2,driven by a comm | vedclass

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(B) The path difference between the waves reaching the detector $D$ from $L_1$ and $L_2$ is given by:$\Delta x = L_2D - L_1D$From the geometry,$L_1D =

Vedclass · Accepted Answer

$330$

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(B) The path difference between the waves reaching the detector $D$ from $L_1$ and $L_2$ is given by:$\Delta x = L_2D - L_1D$From the geometry,$L_1D = 40 \, m$ and the distance between $L_1$ and $L_2$ is $9 \, m$. Thus,$L_2D = \sqrt{40^2 + 9^2} = \sqrt{1600 + 81} = \sqrt{1681} = 41 \, m$.$\Delta x = 41 \, m - 40 \, m = 1 \, m$.For constructive interference (maxima),the condition is $\Delta x = n\lambda$,where $n = 1, 2, 3, ...$.For the first maximum,we take $n = 1$:$1 \, m = 1 \cdot \lambda \implies \lambda = 1 \, m$.The frequency $f$ is given by $f = \frac{v}{\lambda}$,where $v = 330 \, m/s$.$f = \frac{330 \, m/s}{1 \, m} = 330 \, Hz$.

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