Question

Question: Two loud speakers are located 3.35 mapart on an outdoor stage. A listener is18.3m from one andfrom the other. During the sound check, a signal generator drives the two speakers in phase with the same amplitude and frequency. The transmitted frequency is swept through the audible range 20Hz to 20 KHz. (a) What is the lowest frequency f_min1that gives minimum signal (destructive interference) at the listener’s location? By what number must f_min1be multiplied to get(b)The second lowest frequencyf_min2that gives minimum signal and(c)The third lowest frequencyf_min3that gives minimum signal ?(d)What is the lowest frequencyf_min1that gives maximum signal (constructive interference) at the listener’s location ? By what number mustf_min1be multiplied to get(e) the second lowest frequencyf_min2that gives maximum signal and(f) the third lowest frequencyf_min3that gives maximum signal?

Short answer

Answer

1. The lowest frequencyf_min1 that gives minimum signal at the listener’s location is 143 Hz .
2. The number by whichf_min1 must be multiplied to get the second lowest frequencyf_min2that gives minimum signal is 3 .
3. Thethird lowest frequencyf_min3that gives minimum signal is 715 Hz .
4. The lowest frequencyf_max1 that gives the maximum signal at the listener’s location .is 286 Hz
5. The number by which f_max1 must be multiplied to get the second lowest frequency f_max2 that gives maximum signal is 2.
6. The number by whichf_max1 must be multiplied to get the third lowest frequencyf_max3 that gives maximum signal is 3 .

Step-by-step solution

Step 1: Given data

• Velocity of the wave 343 m/s
• Distance from one source is 19.5m.
• Distance from the other source is 18.3m.

Determining the concept

The expression for the velocity of sound in terms of frequency and wavelength is given by,

$v=f\lambda$

Here, v is the velocity ,$\lambda$ is wavelength and f is the frequency.

1. Condition for destructive interference,$f_{min,n}==\frac{\left(2n-1\right)v}{2∆\text{L}},n=1,2,3,\dots$
2. Condition for constructive interference,$f_{max,n}=\frac{nv}{∆L},n=1,2,3,\dots ,n=1,2,3,\dots$

Here is the length.

(a) Determining the lowest frequency fmin1 that gives minimum signal at the listener’s location is fmin1 = 143 Hz

The phase difference is,

$f=\frac{∆L}{\lambda }2\pi$

$∆L$is their path length difference.

Fully constructive interference occurs when and destructive interference occurs when,$f=\left(2n+1\right)\pi$

Therefore, the condition for destructive interference,

$\frac{∆L}{v}f=n-\frac{1}{2}$……. (i)

Where n = 1,2,3,....

From equation, $v=f\lambda$

$\lambda =v/f$

Substitute v/f for$\lambda$ into the equation (i)

$$\begin{gathered} \frac{∆L}{v}f=n-\frac{1}{2} \\ \therefore f=\frac{\left(n-\frac{1}{2}\right)v}{\Delta L} \end{gathered}$$

Now, from the given information, write the path difference as,

$$\begin{gathered} \Delta L=19.5-18.3 \\ =1.2\text{m} \end{gathered}$$

So we can further simplify the equation as,

$$\begin{gathered} f_n=\frac{\left(n-\frac{1}{2}\right)343}{1.2} \\ {f}_n=\left(n-\frac{1}{2}\right)285.833 \end{gathered}$$

…….(ii)

Now, for constructive interference, the condition is,

$f_{max,n}=\frac{mv}{∆L}$

Here n = 1,2,3.....

Substitute 343 m/s for v and$\left(19.5-18.3\right)mfor∆L$ for into the above equation,

role="math" localid="1661326462148" $$\begin{gathered} f_{max,n}=\frac{n\times 343}{\left(19.5-18.3\right)} \\ {f}_{max,n}=\left(286\right)n \end{gathered}$$……. (iii)

From the above equation (ii), the lowest frequency that gives destructive interference is at n = 1,

$$\begin{gathered} f_n=\left(n-\frac{1}{2}\right)285.833 \\ =142.\text{9Hz} \\ \approx 1\text{43Hz} \end{gathered}$$

Therefore the lowest frequency f_min1 that gives minimum signal at the listener’s location is 143 Hz .

(b) Determining the number by which fmin1  must be multiplied to get the second lowest frequency fmin2 that gives minimum signal is 3

From equation (ii), the second lowest frequency that gives destructive interference is at n = 2,

$$\begin{gathered} f_{min,2}=\left(2-\frac{1}{2}\right)286\text{Hz} \\ {f}_{min,2}=\left(\frac{3}{2}\right)286\text{Hz} \\ {f}_{min,2}=429\text{Hz} \end{gathered}$$

So,

$$\begin{gathered} f_{min,2}=3\times 143 \\ {f}_{min,2}=3\times f_{min,1} \end{gathered}$$

Therefore the number by which f_min1 must be multiplied to get the second lowest frequency f_min2 that gives minimum signal is 3.

(c) Determining the third lowest frequency fmin3 that gives minimum signal is fmin3 =715 Hz and the multiplier is  5

From equation (ii), the third lowest frequency that gives destructive interference is at n = 3 ,

$$\begin{gathered} f_{min,3}=\left(3-\frac{1}{2}\right)286\text{Hz} \\ =\left(\frac{5}{2}\right)286\text{Hz} \\ =715Hz \end{gathered}$$

Now write f_min3 in the multiple of

$$\begin{gathered} f_{min,3}=5\times 143 \\ {f}_{min,3}=5\times f_{min,1} \end{gathered}$$

Therefore the third lowest frequency f_min3 that gives minimum signal is 715 Hz .

(d) Determine the lowest frequency  fmax1  that gives maximum signal at the listener’s location

From equation (iii), the lowest frequency that gives constructive interference only when, so that n = 1,

$f_{max,n}=\left(286\right)n$

Substitute 1 for n into the above equation,

$$\begin{gathered} f_{max,1}=\left(286\right)\times 1 \\ =286\text{Hz} \end{gathered}$$

The lowest frequency f_max1 that gives the maximum signal at the listener’s location is286 Hz.

(e) Determining the number by which fmin1   must be multiplied to get the second lowest frequency fmax3  that gives maximum signal

From equation (iii), the second lowest frequency that gives constructive interference is at n = 2,

$$\begin{gathered} f_{max,n}=\left(286\right)n \\ {f}_{max,2}=2\times 286\text{Hz} \\ =572\text{Hz} \\ =2\times f_{max,1}\text{Hz} \end{gathered}$$

Therefore the number by whichf_max1 must be multiplied to get the second lowest frequency f_max2 that gives maximum signal is 2.

(f) Determining the number by which fmin1  must be multiplied to get the third lowest frequency fmax3 that gives maximum signal

From equation (iii), the third lowest frequency that gives constructive interference is at,

$$\begin{gathered} n=3 \\ {f}_{max,n}=\left(286\right)n \\ {f}_{max,3}=3\times 286\text{Hz} \\ =858\text{Hz} \\ =3\times f_{max,1} \end{gathered}$$

So, the multiplication factor to the f_max1is 3 to get thef_max3 .

Therefore the number by which f_max1 must be multiplied to get the third lowest frequency f_max3 that gives maximum signal is 3.