Chapter 17: Q. 68 (page 486)

You are standing 2.5 m directly in front of one of the two loudspeakers shown in FIGURE P17.68. They are 3.0 m apart and both are playing a 686 Hz tone in phase. As you begin to walk directly away from the speaker, at what distances from the speaker do you hear a minimum sound intensity? The room temperature is $20 ° C$ .

Short Answer

Expert verified

The three solutions are

$x_{0} = 17.68 m x_{1} = 5.62 m x_{2} = 2.98 m$

Step by step solution

Description of parameters

Here,

$d = 3.0 m a p a r t f = 686 H z$

The air temperature is $20 ° C$ and sound speed is $343 \frac{m}{s}$

Therefore, the wavelength is $λ = \frac{v}{f} = 0.5 m$

Here, x means the distance from the first loudspeaker and $r_{2}$ means the distance from the second loudspeaker.

Here, the destructive interference condition is $∆ r = (m + \frac{1}{2}) \cdot λ ∆ r = r_{2} - x$

Calculation of the result

As per the theorem of Pythagoras,

$\begin{array}{rcl} r_{2} & = & \sqrt{x^{2} + d^{2}} \\ = & \sqrt{x^{2} + 9} \end{array}$

Thus,

$\begin{array}{rcl} ∆ r & = & \sqrt{x^{2} + 9} - x \\ \sqrt{x^{2} + 9} - x & = & (m + \frac{1}{2}) \cdot λ \\ \sqrt{x^{2} + 9} & = & x - (m + \frac{1}{2}) \cdot λ \end{array}$

Now the square of this equation is

$\begin{array}{rcl} x^{2} + 9 & = & x^{2} + 2 \cdot (m + \frac{1}{2}) \cdot x \cdot λ + {(m + \frac{1}{2})}^{2} \cdot λ^{2} \\ 2 \cdot (m + \frac{1}{2}) \cdot x \cdot λ & = & 9 - {(m + \frac{1}{2})}^{2} \cdot λ^{2} \end{array}$

Therefore, the expression for x is $\frac{\begin{array}{rcl} 9 - {(m + \frac{1}{2})}^{2} \cdot 0.25 \end{array}}{\begin{array}{rcl} m + \frac{1}{2} \end{array}}$