This problem aims to find the **speed** of a bat flying near the **observer** at a **particular frequency.** The concept required to solve this problem is entirely related to **doppler’s effect.**

Suppose that a **sound** or a **wave** of some **frequency** is being produced by a moving source at some **distance** from the **observer,** such that any change in the **frequency** of that **sound** or **wave** generated by that moving **source** with reference to the **observer** is known as **Doppler’s effect.**

In **physics** terms, the **Doppler effect** is the noticeable **change** in the frequency of **sound waves** due to the comparable **motion** between the **source** and the **observer.** We can extrapolate the obvious **frequency** in the **Doppler effect** using the **equation:**

\[f’=\dfrac{(v \pm v_0)}{(v \pm v_s)} f_s\]

Where:

$f’=\text{frequency observed by the observer,}$

$f_s=\text{frequency of the source of the sound,}$

$v=\text{velocity of sound waves or speed of sound,}$

$v_0=\text{velocity of the observer is positive when it is from listener to source,}$

$v_s=\text{velocity of the source is positive when is it from source to listener.}$

This equation can be **changed** in **different situations** relying on the **velocities** of the **observer** or the **source** of the sound waves.

## Expert Answer

When the **sound-generating source** and the **observer** are moving with respect to each other, the **frequency** of the **sound** listened by the **observer** is not equal in **magnitude** to the **source frequency.** For instance, when a **car** comes near you with its **horn blowing,** the **pitch** appears to **decline** as the car **perishes.**

In this problem, we are **requested** to find the **speed** with which the **source** of the **sound** passes by the **observer** so that the **observer** hears a sound of **frequency** $20kHz$. The toughest part is **deciding** the **direction** for each **velocity.**

Since the **source** moves away from the **observer** to make a **frequency** less than its actual **frequency,** a sound of less **frequency** is heard rather than the **actual frequency** from the **source.** Using the **doppler’s equation:**

\[f’=\dfrac{(v \pm v_0)}{(v \pm v_s)} f_s\]

Since the **observer** is **stationary:**

$v_0=0$,

$v_s$ is **positive** as the **source** is **moving away** from the **listener,**

**Plugging** them in:

\[f’=\dfrac{(v + 0)}{(v + v_s)} f_s\]

\[v+v_s=\dfrac{(v\times f_s)}{f’}\]

\[v_s=\dfrac{(v\times f_s)}{f’} – v \]

We have **speed** of **sound** $v = 343 m/s$, the **frequency** of **source** $f_s = 25000 Hz$, and the **frequency** of the **sound** heard by the **listener** $f’ = 20000 Hz$, plugging them in:

\[v_s=\dfrac{((343)\times (25000 ))}{20000 } – 343\]

\[v_s=(343)\times (1.25) – 343 \]

\[v_s=428.75 – 343\]

\[v_s=85.75 m/s \]

## Numerical Result

The **speed** of the **source** is $v_s = 85.75 m/s$.

## Example

**Two** cars are **moving** toward each other at a **speed** of $432 km/h$. If the **frequency** of the **horn blown** by the **first** car is $800Hz$, find the **frequency heard** by the **person** in the **other car.**

The **observer** and the **source** are **moving** toward each other, **therefore,**

\[f’=\dfrac{(v + 0)}{(v – v_s)} f_s \]

**Changing** $432 km/h$ into $m/s$ we get $120 m/s$.

**Substituting** the values:

\[f’=\dfrac{(360 + 120)}{(360 – 120)} 800=1600\space Hz\]