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Sound - Characteristics 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 propagate through a medium as a series of compressions (high pressure) and rarefactions (low pressure).

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Wavelength (λλ): The distance between two consecutive compressions or two consecutive rarefactions. The SI unit is meter (mm).

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

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Time Period (TT): The time taken by two consecutive compressions or rarefactions to cross a fixed point. It is the reciprocal of frequency.

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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; Loudness∝A2\text{Loudness} \propto A^2.

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Speed (vv): The distance traveled by a point on the wave (like a compression) per unit time. Speed remains constant in a given medium at a constant temperature.

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Quality or Timbre: The characteristic that enables us to distinguish one sound from another having the same pitch and loudness.

📐Formulae

ν=1T\nu = \frac{1}{T}

v=λνv = \lambda \nu

v=λTv = \frac{\lambda}{T}

Speed(v)=Distance(s)Time(t)\text{Speed} (v) = \frac{\text{Distance} (s)}{\text{Time} (t)}

💡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} Step 1: Calculate Speed (vv) v=λν=0.35×2000=700 m/sv = \lambda \nu = 0.35 \times 2000 = 700 \text{ m/s} Step 2: Calculate Time (tt) t=dv=1500700≈2.14 st = \frac{d}{v} = \frac{1500}{700} \approx 2.14 \text{ s}

Explanation:

First, convert all units to SI (HzHz, mm). Use the wave equation v=λνv = \lambda \nu to find the velocity of the sound. Finally, use the basic speed-distance-time relation to find the time taken.

Problem 2:

Calculate the total distance traveled by a sound pulse that travels to a wall 750 m750 \text{ m} away and returns. If the original distance was 825 m825 \text{ m} and it was reduced by 75 m75 \text{ m}, show the calculation.

Solution:

The reduction in distance is calculated as: 825−75750\begin{array}{r} 825 \\ - 75 \\ \hline 750 \end{array} Total distance for an echo (to and fro): dtotal=750+750d_{total} = 750 + 750 750+7501500\begin{array}{r} 750 \\ + 750 \\ \hline 1500 \end{array} Total distance = 1500 m1500 \text{ m}.

Explanation:

In echo problems, the sound travels the distance to the obstacle and then travels the same distance back, so the total distance is 2×d2 \times d.