This sheet covers the key formulas and equations from Chapter 10: Sound Waves: Characteristics and Applications. It includes the relation between frequency and time period, the wave equation linking speed, wavelength and frequency, and the formula for echo distance calculation. Each formula is presented with its symbol meanings, SI units, and guidance on when to use it.
Every formula is grouped by topic with its symbols, units, and a short note on the situation where it applies. Original worked examples show how to use each formula step by step. For detailed explanations and derivations, visit the chapter notes page.
Formulas at a Glance
| Purpose (what you are finding) | Formula (MathJax) |
|---|---|
| Time period from frequency (or vice versa) | \( T = \dfrac{1}{\nu} \) |
| Speed of sound using wavelength and time period | \( v = \dfrac{\lambda}{T} \) |
| Speed of sound using wavelength and frequency | \( v = \lambda \nu \) |
| Distance travelled by sound | \( \text{distance} = v \times t \) |
| Distance to reflecting surface (echo) | \( d = \dfrac{v \times t}{2} \) |
All Formulas, Grouped by Topic
Frequency and Time Period
The number of oscillations per second is the frequency \( \nu \). The time taken for one complete oscillation is the time period \( T \).
\[ \nu = \frac{1}{T} \quad \text{or} \quad T = \frac{1}{\nu} \]
Frequency can also be calculated from the number of oscillations \( N \) in a given time \( t \):
\[ \nu = \frac{N}{t} \]
Speed of Sound Wave
The speed \( v \) of a sound wave is the distance a crest (or compression) travels per unit time. It is related to wavelength \( \lambda \) and time period \( T \):
\[ v = \frac{\lambda}{T} \]
Using the frequency \( \nu \), the same relation becomes:
\[ v = \lambda \nu \]
Distance and Echo
When sound travels to a reflecting surface and back, the total distance covered is \( v \times t \). The distance to the surface is half of that:
\[ d = \frac{v \times t}{2} \]
What Each Symbol Means
| Symbol | What it means | SI unit |
|---|---|---|
| \( \nu \) | Frequency of the sound wave | hertz (Hz) or \( \text{s}^{-1} \) |
| \( T \) | Time period of the sound wave | second (s) |
| \( \lambda \) | Wavelength (distance between two consecutive compressions or rarefactions) | metre (m) |
| \( v \) | Speed of sound in the medium | metre per second (m/s) |
| \( t \) | Time taken by sound to travel a certain distance | second (s) |
| \( d \) | Distance from the source to the reflecting surface (or between two points) | metre (m) |
| \( N \) | Number of oscillations (count) | dimensionless (count) |
When to Use Each Formula
- Frequency – time period relation: Use when you know one quantity and need the other. Valid for any periodic wave.
- Frequency from count: Use when you count \( N \) oscillations in a measured time \( t \). Provides the frequency directly.
- Speed of sound (wave equation): Use \( v = \lambda \nu \) when you know any two of speed, wavelength, frequency. The speed depends only on the medium, not on the source.
- Distance travelled by sound: Use \( \text{distance} = v \times t \) for sound travelling in a straight line. This is the basis for echo and sonar calculations.
- Echo distance: Use \( d = \dfrac{v \times t}{2} \) when the time measured is for sound to go to a reflecting surface and come back. The minimum distance for a distinct echo is about 17 m (for \( t = 0.1 \, \text{s} \) and \( v = 340 \, \text{m/s} \)).
Worked Examples
Example 1: Finding frequency from time period
Step 1: A sound wave has a time period of \( 0.005 \, \text{s} \).
What is its frequency?
Step 2: Use the formula \( \nu = \dfrac{1}{T} \).
\[ \nu = \frac{1}{0.005 \, \text{s}} = 200 \, \text{Hz} \]
Final answer: \( \nu = 200 \, \text{Hz} \).
Example 2: Wavelength of a sound wave in air
Step 1: A sound wave of frequency \( 500 \, \text{Hz} \) travels through air at \( 340 \, \text{m/s} \).
Find its wavelength.
Step 2: Use the wave equation \( v = \lambda \nu \), rearranged to \( \lambda = \dfrac{v}{\nu} \).
\[ \lambda = \frac{340 \, \text{m/s}}{500 \, \text{Hz}} = 0.68 \, \text{m} \]
Final answer: \( \lambda = 0.68 \, \text{m} \) (or 68 cm).
Example 3: Echo distance from a cliff
Step 1: A girl shouts near a cliff and hears the echo after \( 0.6 \, \text{s} \).
The speed of sound in air is \( 340 \, \text{m/s} \).
How far is the cliff?
Step 2: The sound travels to the cliff and back.
Use \( d = \dfrac{v \times t}{2} \).
\[ d = \frac{340 \, \text{m/s} \times 0.6 \, \text{s}}{2} = \frac{204 \, \text{m}}{2} = 102 \, \text{m} \]
Final answer: The cliff is \( 102 \, \text{m} \) away.
Common Mistakes to Avoid
| Mistake | Correct rule | How to check your answer |
|---|---|---|
| Treating the echo time as the one-way time | In echo problems, the measured time is for the round trip. Always divide by 2 to get the distance to the reflecting surface. | If your answer is larger than the expected distance, check if you forgot to divide by 2. |
| Confusing frequency \( \nu \) with speed \( v \) | Frequency \( \nu \) (Hz) is the number of oscillations per second; speed \( v \) (m/s) is how fast the wave travels. | Look at the units: if the answer is in Hz, it is frequency; if in m/s, it is speed. |
| Using the wrong unit for time period | Time period \( T \) must be in seconds. If given in milliseconds, convert to seconds before using \( \nu = 1/T \). | Convert all time values to seconds: \( 1 \, \text{ms} = 0.001 \, \text{s} \). |
| Forgetting that speed of sound depends on the medium | Use the correct speed for the given medium (e.g., 340 m/s in air, 1500 m/s in water, 5000 m/s in steel). | Always check the medium stated in the problem and use the table values from the textbook. |
Frequently Asked Questions
What is the formula for the speed of sound?
The speed of sound is given by \( v = \lambda \nu \), where \( \lambda \) is the wavelength (in metres) and \( \nu \) is the frequency (in hertz). The speed depends only on the medium, not on the source frequency.
How do you calculate the distance to an object using echo?
Measure the time \( t \) between the sound emission and the echo. Use \( d = \dfrac{v \times t}{2} \), where \( v \) is the speed of sound in the medium. The division by 2 accounts for the round trip.
What is the minimum distance to hear an echo?
For a distinct echo, the reflected sound must reach the ear at least 0.1 s after the original sound. Using \( v = 340 \, \text{m/s} \), the minimum distance is \( \dfrac{340 \times 0.1}{2} = 17 \, \text{m} \).
How are frequency and time period related?
They are inverses: \( \nu = \dfrac{1}{T} \) and \( T = \dfrac{1}{\nu} \). A higher frequency means a shorter time period.
Reference: NCERT Class 9 Science textbook, chapter Sound Waves: Characteristics and Applications.
Explore Class 9 Physics Formulas
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Related chapters:
- Exploring Mixtures and their Separation notes
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Official source: download the NCERT textbook free from ncert.nic.in.