Home » 國際課程 » IGCSE課程 » Details

Edexcel IGCSE Physics: Double Science 復(fù)習(xí)筆記： 3.1.3 The Wave Equation

Category: IGCSE課程, 國際課程, 教材筆記, 福利干貨 Date: 2022年10月21日下午4:37

Edexcel IGCSE Physics: Double Science 復(fù)習(xí)筆記： 3.1.3 The Wave Equation

The Wave Equation

All waves obey the?wave equation, which is used to calculate the?wave?speed:

Wave-equation

Where:
- v?= wave speed in?metres per second?(m/s)
- f?= frequency in?Hertz?(Hz)
- λ = wavelength in?metres?(m)

The wave speed equation may need to be rearranged, which can be done using this formula triangle:

Wave-speed-triangle-2

Frequency & Time Period

The equation which relates the frequency of a wave to its time period is:

2.1.1.3-Frequency-period-equation

Where:
- T?= time period, measured in?seconds?(s)
- f?= frequency, measured in?Hertz?(Hz)

Worked Example

Visible light has a frequency of about 6 × 10¹⁴?Hz.How long does it take for one complete cycle of visible light to enter our eyes?

Step 1: List the known values

- Frequency, f = 6 × 10¹⁴?Hz

Step 2: State the relationship between frequency and time period?

- This question involves quantities of time and frequency, so the equation which relates?time?period?and?frequency?of a wave is:

7.10-Time-Period-Equation-e1621416317998

Step 3: Substitute the known values in to the equation and calculate the time period?

T =?1 ÷ (6 × 10¹⁴) = 1.67 × 10^-15?s

- Therefore, it takes?1.67 × 10^-15?s?(to 2 decimal places) for one wave of visible light to pass our eyes

Calculations in Different Contexts

The?wave equation?can be applied and rearranged to calculate properties of waves, such as:
- wave speed
- frequency
- time period
The wave equation applies to all types of wave, including?sound?waves?and?electromagnetic?waves

Worked Example

A certain sound wave moves at about 330 m/s and has a time period of 0.0001 seconds.Calculate:

a) The frequency of the sound wave

b) The wavelength of the sound wave

Part (a)

Step 1: List the known quantities

- Time period,?T?= 0.0001 s

Step 2: Write out the equation relating time period and frequency

7.10-Time-Period-Equation-e1621416317998

Step 3: Rearrange for frequency, f, and calculate the answer

f?= 1 ÷?T?= 1 ÷ 0.0001

Frequency,?f?= 10 000 Hz =?1 × 10⁴?Hz

Part (b)

Step 1: List the known quantities

- Wave speed,?v?= 330 m/s
- Frequency,?f?= 1 × 10⁴?Hz

Step 2: Write out the wave speed equation

v?=?f ×?λ

Step 3: Rearrange the equation to calculate the wavelength

λ =?v?÷?f

Step 4: Use the frequency you calculated in part (a) and put the values into the equation

λ = 330 ÷ 1 × 10⁴

Wavelength, λ =?0.033 m

Worked Example

A local radio station broadcasts at a frequency of 200 kHz.The wavelength of these radio waves is 1500 m.Calculate the speed of these radio waves and state an appropriate unit.

Step 1: List the known quantities

- Frequency,?f?= 200 kHz = 200 000 Hz
- Wavelength, λ = 1500 m

Step 2: Write out the wave speed equation

- This question requires wave speed, so state the equation linking wave speed, wavelength and frequency:

v?=?f ×?λ

Step 3: Substitute the known values to calculate the wave speed?

v =?200 000 × 1500 = 300 000 000 = 3 × 10⁸

Step 4: State the unit with the answer

- The units for speed are m/s
- Therefore, the speed of these radio waves is?3 × 10^8?m/s

Exam Tip

When stating equations make sure you use the right letters:

For example, use λ for wavelength, not L or W

If you can’t remember the correct letters, then just state the word equations

Be careful with units: wavelength is usually measured in metres and speed in m/s, but if the wavelength is given in cm you might have to give the speed in cm/s

Likewise, watch out for frequency given in kHz: 1 kHz = 1000 Hz

轉(zhuǎn)載自savemyexams