91影视

Two loudspeakers are in phase. Therefore, the phase difference between the sound waves produced by these is ${蠒}_{鈭�}=0$

Consider the smallest distance between the speakers for which the interference of the sound waves is maximum destructive is $d$.

The frequency of the sound waves $f=686Hz$

At ${20}^{鈭�}C$, the speed of sound in air isrole="math" localid="1650377361718" $v=343m/s.$

For the condition of destructive interference produced by the sound waves emitted by the two speakers, the phase difference between the waves is,

$鈭�蠒=\frac{2蟺鈭�x}{位}...\left(1\right)$

Here, the path-difference between the two waves is, $鈭�x=d$

For destructive interference, the phase difference can be written as,

role="math" localid="1650379152001" $鈭�蠒=2\left(n+\frac{1}{2}\right)蟺...\left(2\right)$where $n=0,1,2...$

Now, for smallest path difference, $n=0$. So, from the above equations 1 and 2,we can write,

role="math" localid="1650379605442" $$\begin{gathered} 2蟺\frac{d}{位}=2\left(0+\frac{1}{2}\right)\mathrm{蟺} \\ 2\mathrm{蟺}\frac{\mathrm{d}}{\mathrm{位}}=\mathrm{蟺} \\ \mathrm{d}=\frac{\mathrm{位}}{2} \\ \mathrm{d}=\frac{\mathrm{v}}{2\mathrm{f}}\left[鈭�\mathrm{位}=\frac{\mathrm{v}}{\mathrm{f}}\right] \\ \mathrm{d}=\frac{343\mathrm{m}/\mathrm{s}}{\left(2脳686\right)\mathrm{Hz}} \\ \mathrm{d}=0.25\mathrm{m} \\ \mathrm{d}=25\mathrm{cm} \end{gathered}$$

When two speakers are out of phase, the phase difference between the sound waves produced by these is ${蠒}_{鈭�}=蟺$

Consider the smallest distance between the speakers for which the interference of the sound waves is maximum constructive is $d\text{'}$

Consider the condition of constructive interference, the path difference between the two waves are

$鈭�蠒=2n蟺...\left(1\right)$where $n=0,1,2....$

Now, the phase difference between the waves can be written in terms of path-difference as,

$$\begin{gathered} 鈭�蠒=\frac{2蟺鈭�x}{位}+{蠒}_{鈭�} \\ 鈭�蠒=\frac{2蟺d\text{'}}{位}+\mathrm{蟺}...\left(2\right) \end{gathered}$$

Consider $n=1,$for the constructive interference pattern, we can write using the two equations 1 and 2 as,

role="math" localid="1650379635522" $$\begin{gathered} \frac{2蟺d\text{'}}{位}+\mathrm{蟺}=2\mathrm{蟺} \\ \frac{2\mathrm{蟺诲}\text{'}}{\mathrm{位}}=\mathrm{蟺} \\ \mathrm{d}\text{'}=\frac{\mathrm{位}}{2} \\ \mathrm{d}\text{'}=\frac{\mathrm{v}}{2\mathrm{f}}\left[鈭�\mathrm{位}=\frac{\mathrm{v}}{\mathrm{f}}\right] \\ \mathrm{d}\text{'}=\frac{343\mathrm{m}/\mathrm{s}}{\left(2脳686\right)\mathrm{Hz}} \\ \mathrm{d}\text{'}=0.25\mathrm{m} \\ \mathrm{d}\text{'}=25\mathrm{cm} \end{gathered}$$