91Ó°ÊÓ

1. The fork with lowest frequency is$f_4=500\text{Hz}$
2. The beat frequencies are$\text{1Hz,2Hz,3Hz,5Hz,7Hz,and8Hz}$
3. There are two sets of answer.

Formula is as follow:

The beat frequency is,

.$f_{beat}=f_1-f_2$

where, f is frequency.

Let,$f_1{f}_{2,}{f}_{3}and{f}_4$be the frequencies of the corresponding fork 1, fork 2, fork 3, and fork 4respectively, so that,$f_1>f_2>f_3>f_4$

.

Then, the beat frequency is,

$f_{beat}=f_1-f_4$

Thus, the highest frequency f₁ of the corresponding fork is given by,

$$\begin{gathered} f_1=f_{beat}+f_4 \\ {f}_1=8+500 \\ {f}_1=508Hz \end{gathered}$$

Similarly, the beat frequency is,

$f_{beat}=f_2-f_4$

Thus, the frequency f₄ of the corresponding fork is given by,

$$\begin{gathered} f_2=f_{beat}+f_4 \\ {f}_2=7+500 \\ {f}_2=507Hz \end{gathered}$$

Similarly, the beat frequency is,

$f_{beat}=f_3-f_4$

Thus, the frequency f₄ of the corresponding fork 3 is given by,

$$\begin{gathered} f_3=f_{beat}+f_4 \\ {f}_3=5+500 \\ {f}_3=505Hz \end{gathered}$$

Now, for second set, also $f_1>f_2>f_3>f_4$and

The beat frequency is,

$f_{beat}=f_1-f_4$

Thus, the highest frequency of the corresponding fork is given by,

$$\begin{gathered} f_1=f_{beat}+f_4 \\ {f}_1=8+500 \\ {f}_1=508H_z \end{gathered}$$

Similarly, the beat frequency is,

$f_{beat}=f_2-f_4$

Thus, the frequency f₂ of the corresponding fork is given by,

role="math" localid="1661435994772" $$\begin{gathered} f_2=f_{beat}+f_4 \\ {f}_2=3+500 \\ {f}_2=503\text{Hz} \end{gathered}$$

Similarly, the beat frequency is,

$f_{beat}=f_3-f_4$

Thus, the frequency of the corresponding fork 3 is given by,

$$\begin{gathered} f_3=f_{beat}+f_4 \\ {f}_3=1+500 \\ {f}_3=501\text{Hz} \end{gathered}$$

Hence, the possible frequencies of other three forks are,$$\begin{gathered} \text{f1=508Hz,f2=507Hz,andf3=505Hz,or,} \\ \text{f1=508Hz,f2=503Hz,andf3=501Hz,} \end{gathered}$$.

Thus, by using the formula for beat frequency,$f_{beat}$this question can be solved.