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 $6.4$ in 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

Hint: In case of destructive interference, we know that all the quantities vary with the path length directly or indirectly and path length varies with the wavelength hence, we will first find the wavelength, and if the sound signal is continuously varying, then a number of frequencies are obtained within the given range.

Formula used:
$\text{wavelength}(\lambda ) = \dfrac{\text{Speed(v)}}{\text{Frequency(f)}}$
$\Delta L = \dfrac{{(2n + 1)\lambda }}{2}$
where $n = $any integer

Complete step by step solution:
Speed of sound in air $v = 320m/s$ (given). The Path difference between the waves coming from both the speakers $\Delta L = 6.4 - 6.0 = 0.4m$.
Now, we know that
$\text{wavelength}(\lambda ) = \dfrac{\text{Speed(v)}}{\text{Frequency(f)}}$
i.e., $\lambda = \dfrac{{320}}{f}$ … (1)
Also, we know that in case of destructive interference
$\Delta L = \dfrac{{(2n + 1)\lambda }}{2}$
where $n = $any integer
Substituting $\Delta L = 0.4m$ and value of $\lambda $ from eq. (1) in the above expression, we get
$0.4 = \dfrac{{(2n + 1)}}{2}.\dfrac{{320}}{f}$
Thus, $f = (2n + 1)400Hz$
At $n = 1,f = (2(1) + 1)400 = 1200Hz$
At $n = 2,f = (2(2) + 1)400 = 2000Hz$
At $n = 3,f = (2(3) + 1)400 = 2800Hz$
At $n = 4,f = (2(4) + 1)400 = 3600Hz$
At $n = 5,f = (2(5) + 1)400 = 4400Hz$

Hence, The frequencies that cause destructive interference are $1200Hz$ , $2000Hz$ , $2800Hz$ , $3600Hz$ , $4400Hz$ within the specified range of $500Hz - 5000Hz$ on different values of $n$.

Note:Since this is a problem related to destructive interference hence, given conditions are to be analyzed very carefully and quantities that are required to calculate path length must be identified on a prior basis as it gives a better understanding of the problem. Units must be put after each end result.