Question

Two identical tuning forks can oscillate at $\mathbf{440}\mathbf{\text{\hspace{0.17em}Hz}}$. A person is located somewhere on the line between them. Calculate the beat frequency as measured by this individual if (a) she is standing still and the tuning forks move in the same direction along the line at,$\mathbf{3}\mathbf{.00}\mathbf{\text{\hspace{0.17em}m}}\mathbf{/}\mathbf{\text{s}}$ (b) the tuning forks are stationary and the listener moves along the line at $\mathbf{3}\mathbf{.00}\mathbf{\text{\hspace{0.17em}m}}\mathbf{/}\mathbf{\text{s}}$.

Short answer

1. The beat frequency is $7.70\text{\hspace{0.17em}Hz}$ when the person is standing still and tuning forks are moving in a direction with $3.00\text{\hspace{0.17em}m}/\text{s}$.
2. The beat frequency is $7.70\text{\hspace{0.17em}Hz}$, when the person moves along a line at $3.00\text{\hspace{0.17em}m}/\text{s}$ and the tuning forks are stationary.

Step-by-step solution

The given data

1. Frequency of the tuning fork $=440\text{\hspace{0.17em}Hz}$
2. In part a Velocity of both the tuning forks$V_s=3.00\text{\hspace{0.17em}m}/\text{s}$and Velocity of person$V_D=0\text{\hspace{0.17em}m}/\text{s}$
3. In part b Velocity of both the tuning forks,$V_s=0.00\text{\hspace{0.17em}m}/\text{s}$and Velocity of person,$V_D=3.00\text{\hspace{0.17em}m}/\text{s}$
4. Velocity of the speed of sound, $V=343\text{\hspace{0.17em}m}/\text{s}$

Understanding the concept of the Doppler Effect

As the person is stationary, and the tuning forks are moving along the line, it causes interference. Using the Doppler Effect, we can find the frequency of the sound created by both the tuning forks heard by the person. As we get the respective frequencies we take the difference between them to get the beat frequency. We use the same procedure for part b).

Formula:

The frequency received by the observer or the source according to Doppler’s Effect,

$\mathit{f}\mathbf{\text{'}}\mathbf{=}\mathit{f}\left(\frac{V\pm V_D}{V\pm V_s}\right)$ …(i)

The beat frequency,

${\mathit{f}}_{\mathbf{b}\mathbf{e}\mathbf{a}\mathbf{t}}\mathbf{=}{\mathit{f}}_{\mathbf{1}\mathbf{s}\mathbf{t}\mathbf{}\mathbf{t}\mathbf{u}\mathbf{n}\mathbf{n}\mathbf{i}\mathbf{n}\mathbf{g}\mathbf{}\mathbf{f}\mathbf{o}\mathbf{r}\mathbf{k}}^{\mathbf{\text{'}}}\mathbf{-}{\mathit{f}}_{\mathbf{2}\mathbf{n}\mathbf{d}\mathbf{}\mathbf{t}\mathbf{u}\mathbf{n}\mathbf{i}\mathbf{n}\mathbf{g}\mathbf{}\mathbf{f}\mathbf{o}\mathbf{r}\mathbf{k}}^{\mathbf{\text{'}}}$ …(ii)

a) Calculation of the beat frequency when person is still and tuning forks are moving

Let$f_1$and$f_2$be the frequency observed by the person due tothe 1stand 2^nd tuning fork, respectively. Whenthepersonis standing still and the tuning forks are moving in a direction with. From the equation (i), we can get the frequencies of both tuning forks as:

Frequency of tuning fork 1,

$\begin{array}{l}{f}_1=440\text{\hspace{0.17em}Hz}\times \left(\frac{343\text{\hspace{0.17em}m}/\text{s}+0}{343\text{\hspace{0.17em}m}/\text{s}-3.00\text{\hspace{0.17em}m}/\text{s}}\right)\\ f_1=443.9\text{\hspace{0.17em}Hz}\end{array}$

Frequency of tuning fork 2,

$\begin{array}{l}{f}_1=440\text{\hspace{0.17em}Hz}\times \left(\frac{343\text{\hspace{0.17em}m}/\text{s}+0}{343\text{\hspace{0.17em}m}/\text{s}+3.00\text{\hspace{0.17em}m}/\text{s}}\right)\\ f_1=436.2\text{\hspace{0.17em}Hz}\end{array}$

So, the beat frequency using equation (ii) is given as:

$\begin{array}{l}{f}_{beat}=443.9\text{\hspace{0.17em}Hz}-436.2\text{\hspace{0.17em}Hz}\\ f_{beat}=7.70\text{\hspace{0.17em}Hz}\end{array}$

Hence, the value of the required frequency is$7.70\text{\hspace{0.17em}Hz}$ .

b) Calculation of the beat frequency when person is moving and tuning forks are stationary

Let$f_1^{\text{'}}$and$f_2^{\text{'}}$be the frequency observed by the person due to 1^st and 2^nd tuning fork respectively,

When the person moves along a line at, andthetuning forks are stationary.From the equation (i), we can get the frequencies of both tuning forks as:

Frequency of tuning fork 1,

$\begin{array}{l}{f}_1=440\text{\hspace{0.17em}Hz}\times \left(\frac{343\text{\hspace{0.17em}m}/\text{s}-3.00\text{\hspace{0.17em}m}/\text{s}}{343\text{\hspace{0.17em}m}/\text{s}+0}\right)\\ f_1=436.1\text{\hspace{0.17em}Hz}\end{array}$

Frequency of tuning fork 2,

$\begin{array}{l}{f}_1=440\text{\hspace{0.17em}Hz}\times \left(\frac{343\text{\hspace{0.17em}m}/\text{s}+3.00\text{\hspace{0.17em}m}/\text{s}}{343\text{\hspace{0.17em}m}/\text{s}+0}\right)\\ f_1=443.8\text{\hspace{0.17em}Hz}\end{array}$

So, the beat frequency using equation (ii) is given as:

$\begin{array}{l}{f}_{beat}=|436.1\text{\hspace{0.17em}Hz}-443.8\text{\hspace{0.17em}Hz}|\\ f_{beat}=7.70\text{\hspace{0.17em}Hz}\end{array}$

Hence, the value of the required frequency is $7.70\text{\hspace{0.17em}Hz}$.