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Sound waves travel at 350 m/s through a warm air and at 3500 m/s through brass. The wavelength of a 700 Hz acoustic wave as it enters brass from warm air 
1. decrease by a factor of 10 
2. increase by a factor of 20
3. increase by a factor of 10 
4. decrease by a factor of 20

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Correct Answer - Option 3 : increase by a factor of 10 

CONCEPT:

  • wave can be described as a disturbance that travels through a medium from one location to another location.
  • Properties of wave:
    • Frequency: It is a measure of how many waves pass a point in a certain amount of time.
      • The higher the frequency, the closer the waves are together and the greater the energy carried by the waves will be.
    • Amplitude: Amplitude is a measure of the distance between a line through the middle of a wave and a crest or trough.
      • The greater the force that produces a wave, the greater the amplitude of the wave and the greater the energy carried by the wave.
      • The highest point of a transverse wave is the crest and the lowest point is called a trough.
      • Sounds with greater amplitude will be louder; light wither greater amplitude will be brighter.
    • Wavelength : Wavelength is a measure of the distance from the crest on one wave to the crest on the very next wave.
      • shorter wavelengths are influenced by the frequency.
      • A higher frequency causes a shorter wavelength and greater energy.
    • Speed : Speed is a measure of the distance a wave travels in an amount of time.
      • The speed of a wave is determined by the type of wave and the nature of the medium.
      • As a wave enters a different medium, the wave's speed changes. Waves travel at different speeds in different media. 

When a wave travels from one medium to another, its frequency remains same where as velocity and wavelength changes.

The speed, wavelength anf frequency are related by v = ν λ;

Calculation:

Given vwarm-air = 350 m/s, vbrass = 3500 m/s; ν = 700 Hz;

Since frequency remains the same,

⇒ \(\frac {v_{warm-air}}{v_{brass}} = \frac {λ_{warm-air}}{λ_{brass}}\)

⇒ λbrass = λwarm-air;

∴ The wavelength increases by a factor of 10 when it goes to brass from warm-air.  

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