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Sound - Graphical Representation of a Sound Wave

Grade 9CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Sound waves are longitudinal waves that can be graphically represented as a sine wave, plotting the changes in density or pressure of the medium against distance or time.

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The regions of maximum density and pressure are called Compressions, represented by the 'crests' or peaks in the graph.

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The regions of minimum density and pressure are called Rarefactions, represented by the 'troughs' or valleys in the graph.

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Wavelength (λλ): The distance between two consecutive compressions (crests) or two consecutive rarefactions (troughs). Its SI unit is the metre (mm).

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Frequency (νν): The number of complete oscillations per unit time. It determines the pitch of the sound. Its SI unit is Hertz (HzHz).

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Time Period (TT): The time taken for one complete oscillation in the density of the medium. Its SI unit is second (ss).

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Amplitude (AA): The magnitude of the maximum disturbance in the medium on either side of the mean value. It determines the loudness of the sound (Loudness ∝A2\propto A^2).

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Speed (vv): The distance travelled by a point on the wave (such as a compression) per unit time.

📐Formulae

ν=1T\nu = \frac{1}{T}

Speed (v)=Wavelength (λ)Time Period (T)\text{Speed } (v) = \frac{\text{Wavelength } (\lambda)}{\text{Time Period } (T)}

v=λνv = \lambda \nu

💡Examples

Problem 1:

A sound wave has a frequency of 2 kHz2\text{ kHz} and a wavelength of 35 cm35\text{ cm}. How long will it take to travel 1.5 km1.5\text{ km}?

Solution:

Given: Frequency ν=2 kHz=2000 Hz\nu = 2\text{ kHz} = 2000\text{ Hz}, Wavelength λ=35 cm=0.35 m\lambda = 35\text{ cm} = 0.35\text{ m}, Distance d=1.5 km=1500 md = 1.5\text{ km} = 1500\text{ m}.

First, calculate the speed of the wave: v=λνv = \lambda \nu v=0.35×2000=700 m/sv = 0.35 \times 2000 = 700\text{ m/s}

Now, calculate the time: Time (t)=Distance (d)Speed (v)\text{Time } (t) = \frac{\text{Distance } (d)}{\text{Speed } (v)} t=1500700≈2.14 st = \frac{1500}{700} \approx 2.14\text{ s}

Explanation:

We first converted all units to SI. Using the relationship between speed, wavelength, and frequency, we found the velocity. Finally, we used the basic speed-distance-time formula to find the time taken.

Problem 2:

If the time period of a vibrating body is 0.02 s0.02\text{ s}, calculate the frequency of the sound wave produced.

Solution:

Given: Time Period T=0.02 sT = 0.02\text{ s}.

Frequency is the reciprocal of the time period: ν=1T\nu = \frac{1}{T} ν=10.02\nu = \frac{1}{0.02} ν=50 Hz\nu = 50\text{ Hz}

Explanation:

The frequency of a wave is defined as the number of oscillations per second, which is mathematically the inverse of the time period.