Pitch and Loudness

Sound travels as a longitudinal wave where particles vibrate back and forth as they pass energy on to neighbouring particles. It is useful to visualise sound as a transverse wave when exploring the properties of sound. Two common properties are pitch and loudness. Pitch is often referred to as how high or low a note is, whilst loudness refers to well, how loud it is or how much sound energy it has.

Pitch and loudness can be analysed by looking at the amplitude and wavelength of a transverse wave that represents the sound wave. An oscilloscope can be used to observe sound wave patterns as transverse waves. An audio generator can be used to generate a sound that is played through a speaker. A microphone connected to an oscilloscope can be used to detect the sound wave, whilst the oscilloscope will display the sound wave as a transverse wave pattern on a screen.

The pitch of a sound is related to frequency, which is related to the wavelength of a wave. The higher the frequency (shorter wavelength), the higher the pitch. The loudness of a sound wave is related to the amplitude. A bigger amplitude results in a louder sound.

The diagram below illustrates the properties loudness and pitch in terms of amplitude and frequency:

Example 1:

Using the wave below, draw two more sound waves:

a) same pitch, softer

b) higher pitch, louder

 

a) is in orange, b) is in blue

Example 2:

a) A sound wave has a frequency of 1000Hz and an amplitude of 5cm. If the sound is altered to have a pitch twice as high and it is also twice as loud, how does this impact the frequency and amplitude?

Frequency = 2000Hz

Amplitude = 10cm

b) What is the wavelength of the new sound wave?

Using v=f\lambda

v=340{ m }/{ s }

f=2000Hz

\lambda =\cfrac { v }{ f }

\lambda =\cfrac { 340 }{ 2000 }

\lambda =0.17m

 

 

Horizontal Blocks in Contact

Horizontal blocks in contact is a classic problem that considers both Newton’s 2nd and 3rd laws. Consider two objects A and B, with masses of 5kg and 3kg respectively, that are being pushed with a force of 40N to the right as shown below:

Newtons 3rd Law predicts that A exerts a force on B and B exerts an equal but opposite force on A:

{ F }_{ AB }=-F_{ BA }

We can consider a few aspects of this problem:

If the total force acting on both objects is 40N and they have a combined mass of 8kg, according to Newton’s 2nd law the acceleration of both objects:

a=\cfrac { F }{ m }

a=\cfrac { 40 }{ 8 }

a=5\cfrac { m }{ { s }^{ 2 } }

We can now determine the net force acting on each object:

Object A: F=ma

F=5\times 5

F=25\quad N

Object B: F=ma

F=3\times 5

F=15\quad N

Based on Newtons 3rd we can see that the reaction force of B on A; F_{ BA }=15N

We can now observe the net force acting on A, F_{ net }=40-15=25N which agrees with our working above. 

Solving these problems requires an understanding of the forces analysed above.

Example 1:

In the diagram below, box A has a mass of 8kg and box B has a mass of 5kg. A force of 78N acts on the boxes to the right. Calculate the following:

a) the acceleration of the boxes

b) the force acting on box A and B

c) the reaction force of B on A

a) Using Newtons 2nd law: F=ma

a=\cfrac { F }{ m }

a=\cfrac { 78 }{ (8+5) }

a=6\quad \cfrac { m }{ { s }^{ 2 } }

b) Again, using F=ma and the answer to part a:

Box A: F=ma

F=8\times 6

F=48\quad N right

Box B: F=ma

F=5\times 6

F=30\quad N right

c) The reaction force of B on A is equal and opposite to the force of A on B:

F_{ BA }=30N left