91影视

1. Form of sound wave is $s\left(x,t\right)=s_m\cos \left(kx-蝇t+蠁\right)$
2. Speed of the sound in air, $v=343\text{m/s}$
3. For molecule A,$x_A=2.000\text{misatmaximumpositivedisplacement,}{S}_A=6\text{nm}$
4. For molecule B, $x_B=2.070\text{misatpositivedisplacement,}{S}_B=\text{2nm}$
5. Maximum positive displacement i.e. amplitude, $S_m=\text{6nm}$

By using the values of displacements for molecule A and B, in the given form of sound wave, two equations are found. By subtracting and solving them we can find the frequency.

The displacement in sound wave is given as-

$s\left(x,t\right)=s_m\cos \left(kx-蝇t+蠁\right)$

where, k is spring constant, s, x are displacements,tis time and w is angular frequency.

The form of a sound wave is,

$s=s_m\cos \left(kx-蝇t+蠁\right)$

Applying it for molecule A,

role="math" localid="1661495477076" $$\begin{gathered} s_A=s_m\cos \left(k{x}_A-蝇t+蠁\right) \\ {\cos }^{-1}\left(\frac{S_a}{S_m}\right)=k{x}_A-蝇t+蠁鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�\left(1\right) \end{gathered}$$

Similarly for molecule B,

$$\begin{gathered} s_B=s_m\cos \left(k{x}_B-蝇t+蠁\right) \\ {\cos }^{-1}\left(\frac{S_b}{S_m}\right)=k{x}_B-蝇t+蠁鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�\left(2\right) \end{gathered}$$

Subtracting equation 1 from 2

${\cos }^{-1}\left(\frac{S_b}{S_m}\right)-{\cos }^{-1}\left(\frac{S_a}{S_m}\right)=k(x_B-x_A)鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�\left(2\right)$

For the sound wave amplitude, $S_m=6\text{nm}$

For molecule A, $x_A=2.0\text{00m,}{S}_A=6\text{nm}$

For molecule B,$x_B=2.\text{070m,}{S}_B=2\text{nm}$

Using all these values in equation 3,

${\cos }^{-1}\left(\frac{2\text{nm}}{6\text{nm}}\right)-{\cos }^{-1}\left(\frac{6\text{nm}}{6\text{nm}}\right)=k(2.070\text{m}-2.000\text{m})鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�\left(2\right)$

$$\begin{gathered} k\left(0.07\text{m}\right)=Co{s}^{-1}\left(\frac{1}{3}\right)-Co{s}^{-1}\left(1\right) \\ k\left(0.07\text{m}\right)=1.23-0 \\ k=\frac{1.23}{0.07\text{m}} \\ k=17.57{\text{m}}^{-1}鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�\text{- - - - -}\left(4\right) \end{gathered}$$

Now, angular frequency is given by,

$蝇=kv$

And frequency is given by,

$$\begin{gathered} f=\frac{蝇}{2蟺} \\ =\frac{kv}{2蟺} \end{gathered}$$

Usingthevalue of velocity and wave number,

$$\begin{gathered} f=\frac{kv}{2蟺} \\ f=\frac{17.57\text{m}脳343m/s}{2蟺} \\ f鈮�96\text{0Hz} \end{gathered}$$

Hence, frequency of the sound wave is 960 Hz .