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Introduction to Acoustics

On this page you will find the answers to the exercises in the booklet 'Introduction to Acoustics', Fifth Edition by Mark Huckvale.

Chapter One

1. Fill in the blanks:

  1. The frequency of a periodic event is measured in hertz.
  2. An event that repeats 1000 times per second has a frequency of 1kHz.
  3. The density of air is about 1kgm−3.
  4. An event that repeats 10,000 times per second has a period of 100µs.
  5. 5cm is 5x10−2m = 0.05m.

2. Express the following in scientific notation to three significant figures:

  1. 61743 = 6.17 x 104
  2. 20000 = 2.00 x 104
  3. 0.02 = 2.00 x 10−2
  4. 1/500 = 2.00 x 10−3
  5. 1/0.05 = 2.00 x 101
  6. 2556 x 10−4 = 2.56 x 10−1
  7. 0.0564 x 10 = 5.64 x 103

3. Simplify the following:

  1. 104 x 105 = 109
  2. 104 x 10-3 = 101
  3. 10−6 x 103 x 10 = 10−2
  4. 101 ÷ 10−3 = 104

Chapter Two

1. Given the following (approximate) logarithms to base 10:

    log(2) = 0.3; log(3) = 0.48; log(4) = 0.6; log(5) 0.7

Find approximate solutions to the following sums without using a calculator:

  1. log(100) = 2
  2. log(0.01) = −2
  3. 100 = 1
  4. 100.3 = 2
  5. log(6) = 0.3 + 0.48 = 0.78
  6. log(0.25) = −log(4) = −0.6
  7. log(2/3) = 0.3 − 0.48 = −0.18
  8. 101.7 = 101 x 100.7 = 50

2. The following measurements were made from a straight line graph:

    When X is 0, Y = 12

    Change in Y for 4 units along the X axis = 3.2

  1. What is the equation of the line?

    Y = 0.8X + 12

  2. What is the value of Y when X is 20?

    28

  3. What is the value of X when Y is 20?

    10

Chapter Three

1. How much chocolate would I need to spread over a 1m2 surface to generate a pressure underneath of 1Pa? (assume g=10ms−2) If I spread that amount of chocolate over 1/20 of the area, how does the pressure change?

  • Pressure = Force / Area
  • Force = Mass x Acceleration
  • Therefore: Mass = Pressure x Area / Acceleration
  • Mass = 1 x 1 / 10 = 0.1kg
  • Reducing the area by 20 increases the pressure by 20 for a fixed size force.

2. Winding a grandfather clock involves lifting a weight of 250g up a distance of 0.8m. Taking the acceleration due to gravity (g) to be 10ms−2, how much more energy does the fully wound clock have over the fully unwound one?

  • Energy = Force x Distance
  • Force = Mass x Acceleration
  • Therefore: Energy = Mass x Acceleration x Distance
  • Energy = 0.25 x 0.8 x 10 = 2J

3. If the intensity of a sound at 2m from the source is 10Wm−2 , what would be the intensity at 6m from the source if the energy was radiated equally in all directions? (Hint: the total power at 2m is distributed over the surface of a sphere of radius 2m, but the total power at 6m is distributed over the surface of a sphere of radius 6m).

  • Intensity = Power / Area
  • Power = Intensity x Area
  • Call the area of the sphere at 2m = A
  • Then Power = 10W x A
  • When sphere grows to 6m, radius increases by x3
  • But Area increases by x9 = 9 x A
  • Therefore Intensity at 6m = (10W x A) / (9 x A) = 1.1W
  • Intensity falls as the square of the distance to the source: this is called the inverse square law.


© 2003 Mark Huckvale University College London