**– In milliseconds, how long does it take for a sound wave to vibrate at a frequency at 784 Hz, or the pitch of the G5 on a piano? **

**– What’s the wavelength of an acoustic source one octave greater than the uppermost note? **

The main objective of this question is to calculate the **time** required for a sound wave to** vibrate** at a given frequency and **the wavelength **of an **acoustic source**.

This question uses the concept of **wavelength**, **frequency** and** speed of the wave**. The distance among **identical locations** in adjacent **phases** of a waveform **pattern** carried in** air** or via a** wire** is defined as its **wavelength** and **frequency** is defined as **reciprocal** of **time period**.

## Expert Answer

a) We **know** that:

\[ \space v \space = \space f \space . \space \lambda \]

**And**:

\[ \space T \space = \space \frac{1}{f} \]

Given **that**:

\[ \space f_1 \space = \space 784 Hz \]

\[ \space v \space = \space 344 \frac{m}{s} \]

By **putting values**, we get:

\[ \space 344 \frac {m}{s} \space = \space (784 s^{-1}) \lambda_1 \]

By **simplifying**, we get:

\[ \space \lambda_1 \space = \space 0.439 m \]

The **time period** is given as:

\[ \space T_1 \space = \space \frac{1}{784} \]

\[ \space T_1 \space = \space 1.28 \space \times \space 10^{-3} \]

\[ \space T_1 \space = \space 1.28 \]

b) The **wavelength** of an acoustic source one **octave greater** than the uppermost note is **calculated** as:

\[ \space f_2 \space = \space 2 \space \times \space f_1 \]

By **putting** values, we get:

\[ \space = \space 2 \space \times \space 784 \]

\[ \space = \space 1568 hz \]

**Now**:

\[ \space 344 \frac {m}{s} \space = \space (1568 s^{-1}) \lambda_2 \]

By **simplifying**, we get:

\[ \space \lambda_2 \space = \space 0.219 m \]

## Numerical Results

The time required for a sound wave to vibrate at a given frequency is:

\[ \space T_1 \space = \space 1.28 \]

The wavelength is:

\[ \space \lambda_2 \space = \space 0.219 m \]

## Example

In **milliseconds**, how long does it take for a **sound wave** to vibrate at a **frequency** at $ 800 Hz $** when** the sound speed is 344 \frac{m}{s} at 20 C \{circ} in air. **What’s** the wavelength** of an **acoustic source** one octave greater **than **the **uppermost** note? **

We **know** that:

\[ \space v \space = \space f \space . \space \lambda \]

**And**:

\[ \space T \space = \space \frac{1}{f} \]

Given **that**:

\[ \space f_1 \space = \space 800 Hz \]

\[ \space v \space = \space 344 \frac{m}{s} \]

By **putting values**, we get:

\[ \space 344 \frac {m}{s} \space = \space (800 s^{-1}) \lambda_1 \]

By **simplifying**, we get:

\[ \space \lambda_1 \space = \space 0.43 m \]

The **time period** is given as:

\[ \space T_1 \space = \space \frac{1}{784} \]

\[ \space T_1 \space = \space 1.28 \space \times \space 10^{-3} \]

\[ \space T_1 \space = \space 1.28 \]

**Now t**he **wavelength** of an acoustic source one **octave greater** than the uppermost note is **calculated** as:

\[ \space f_2 \space = \space 2 \space \times \space f_1 \]

By **putting** values, we get:

\[ \space = \space 2 \space \times \space 784 \]

\[ \space = \space 1568 hz \]

**Now**:

\[ \space 344 \frac {m}{s} \space = \space (1568 s^{-1}) \lambda_2 \]

By **simplifying**, we get:

\[ \space \lambda_2 \space = \space 0.219 m \]