A person standing at a distance of 6, m from | vedclass

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(A) The person receives sound via two paths: the direct path of length $d_1 = 6\, m$ and the reflected path of length $d_2 = 5\, m + 5\, m = 10\, m$.T

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(A) The person receives sound via two paths: the direct path of length $d_1 = 6\, m$ and the reflected path of length $d_2 = 5\, m + 5\, m = 10\, m$.The path difference between the two waves is $\Delta x = d_2 - d_1 = 10\, m - 6\, m = 4\, m$.For constructive interference (maximum sound intensity),the path difference must be an integral multiple of the wavelength $\lambda$:$\Delta x = n\lambda$,where $n = 1, 2, 3, \dots$Substituting the values,we get $4 = n\lambda$,which implies $\lambda = \frac{4}{n}$.To find the maximum wavelength,we must choose the smallest positive integer value for $n$,which is $n = 1$.Therefore,$\lambda_{\max} = \frac{4}{1} = 4\, m$.

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