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(C) The distance from the first speaker to point $P$ is $d_1 = 4 \,m$. The distance from the second speaker to point $P$ is $d_2 = \sqrt{3^2 + 4^2} =

Vedclass · Accepted Answer

$\lambda_1=2 \,m$ and $\lambda_2=1 \,m$

Vedclass · Answer

(C) The distance from the first speaker to point $P$ is $d_1 = 4 \,m$. The distance from the second speaker to point $P$ is $d_2 = \sqrt{3^2 + 4^2} = 5 \,m$.The path difference between the two sound waves at point $P$ is $\Delta L = d_2 - d_1 = 5 \,m - 4 \,m = 1 \,m$.For the sound to be least intense (destructive interference),the path difference must be an odd multiple of half the wavelength:$\Delta L = (2n + 1) \frac{\lambda_1}{2}$,where $n = 0, 1, 2, \dots$For $n = 0$,$1 \,m = \frac{\lambda_1}{2} \Rightarrow \lambda_1 = 2 \,m$.For the sound to be most intense (constructive interference),the path difference must be an integer multiple of the wavelength:$\Delta L = n \lambda_2$,where $n = 1, 2, 3, \dots$For $n = 1$,$1 \,m = \lambda_2 \Rightarrow \lambda_2 = 1 \,m$.Thus,the possible values are $\lambda_1 = 2 \,m$ and $\lambda_2 = 1 \,m$. The correct option is $(c)$.

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