LEARNATHON

III

Competition for grade 6 to 10 students! Learn, solve tests and earn prizes!

Learn more### Theory:

The distance travelled by a sound wave per unit time as it propagates through an elastic medium is known as the speed of sound.

$\mathit{Speed}(v)=\frac{\mathit{Distance}}{\mathit{Time}}$

If one wavelength ($\mathrm{\lambda}$) represents the distance travelled by one wave, and one time period (\(T\)) represents the time taken for this propagation, then

$\mathit{Speed}(v)\phantom{\rule{0.147em}{0ex}}=\frac{\mathit{One}\phantom{\rule{0.147em}{0ex}}\mathit{wavelength}(\mathrm{\lambda})}{\mathit{One}\phantom{\rule{0.147em}{0ex}}\mathit{time}\phantom{\rule{0.147em}{0ex}}\mathit{period}(T)}$

And, we know

$T=\frac{1}{v}$

By applying this in speed formula, we get

$v=n\mathrm{\lambda}$

Under the same physical conditions, the speed of sound in a given medium remains nearly constant for all frequencies.

Let us solve following example for better understanding.

**Example**:

A sound wave has a frequency of \(2\) \(kHz\) and a wavelength of \(15\) \(cm\). How much time will it take to travel \(1.5\) \(km\)?

**Given data**:

Frequency \(=\) \(2\) \(kHz\) \(=\) \(2000\) \(Hz\)

Wavelength \(=\) \(15\) \(cm\) \(=\) \(0.15\) \(m\)

Distance \(=\) \(1.5\) \(km\) \(=\) \(1500\) \(m\)

**To find**: Time period

**Formula**:$\mathit{Time}=\frac{\mathit{Distance}}{\mathit{Speed}}$

We don't know the value of speed,

$v=n\mathrm{\lambda}$

$\begin{array}{l}v=2000\times 0.15\\ =300\phantom{\rule{0.147em}{0ex}}m/s\end{array}$

Now, apply in the value of speed in time formula

$\begin{array}{l}\mathit{Time}=\frac{1500}{300}\\ =5\phantom{\rule{0.147em}{0ex}}s\end{array}$

The sound will take \(5\) \(s\) to travel a distance of \(1.5\) \(km\).