Question

Two audio speakers are kept some distance apart and are driven by the same amplifier system. A person is sitting at a place $6.0m$ from one of the speakers and $6.4m$ from the other. If the sound signal is continuously varied from $500Hz$ to $5000Hz$, what are the frequencies for which there is a destructive interference at the place of the listener? Speed of sound in air $=320m/s$.

Answer 1

Step 1: Given Data

Position of the first speaker from the listener $=6.0m$

Position of the second speaker from the listener $=6.4m$

Speed of sound in air $=320m/s$

Range of frequency $=500Hz$ to $5000Hz$

Let the frequency be $f$.

Let the wavelength be $\lambda$.

Let $n$ be an integer such that $n=1,2,3,.....\infty$.

Step 2: Formula Used

For destructive interference,

$△L=\frac{2n+1}{2}\lambda$

Step 3: Establish the Frequency

Path difference can be given as,

$△L=6.4-6.0=0.4m$

We know that wavelength is given as,

$\lambda =\frac{v}{f}=\frac{320}{f}$

For destructive interference,

$$\begin{gathered} △L=\frac{2n+1}{2}\lambda \\ \Rightarrow 0.4=\frac{2n+1}{2}.\frac{320}{f} \\ \Rightarrow f=\frac{2n+1}{2}.\frac{320}{0.4} \\ \Rightarrow f=\left(2n+1\right)400Hz \end{gathered}$$

Step 4: Calculate the Frequencies

For $n=1$,

$f=\left(2+1\right)400=1200Hz$

For $n=2$,

$f=\left(4+1\right)400=2000Hz$

For $n=3$,

$f=\left(6+1\right)400=2800Hz$

For $n=4$,

$f=\left(8+1\right)400=3600Hz$

For $n=5$,

$f=\left(10+1\right)400=4400Hz$

From $n=6,7,8.....\infty$, the value of frequency extends the given frequency range.

Hence, the frequencies for which there is a destructive interference at the place of the listener are $1200Hz,2000Hz,2800Hz,3600Hz$ and $4400Hz$.