Beats - Learn - ScienceFlip

Beats

When two waves superimpose on each other they can create a new wave or cancel each other out. This is known as the superposition of waves. When two sound waves superimpose on each other they generate points of constructive and destructive interference. These points correspond to points of higher and lower volume.

When two sound waves of equal volume but different wavelength (frequency) superimpose the points of constructive interference (higher volume) and destructive interference (lower volume) form a regular pulse like sound called a beat. This beat occurs at a regular rate known as the beat frequency.

The beat frequency can be determined using:

${ f }_{ beat }=|{ f }_{ 2 }-{ f }_{ 1 }|$

where:

${ f }_{ 1 }$ and ${ f }_{ 2 }$ are the two source frequencies. The parallel vertical lines (||) represent the absolute value of the calculation, that is, a positive value for the beat frequency.

Demonstration: Open the following tone generator in two separate tabs. Play different tones on each tab (2 or 3 Hz difference is good to begin with. Observe the beat frequency. Play around with different tones and different beat frequencies.

Example 1:

Two guitars play a note each with frequencies of 270Hz and 266Hz respectively. What is the beat frequency heard by a listener?

Answer:

${ f }_{ 1 }$ = 270Hz

${ f }_{ 2 }$ = 266Hz

${ f }_{ beat }=|266-270|$

${ f }_{ beat }=|-4|$

${ f }_{ beat }=4$ Hz

Force-Time Graphs

A force-time graph will have force (N) on the vertical axis and time (s) on the horizontal. The area under a force-time graph is $Ft$ , which is the impulse the object will experience during that time interval ( $I=Ft$ ). Recall that impulse is equal to the change in momentum of an object.

Example 1:

The force acting on an object over a 15 second interval is shown below:

a) Determine the force acting on the object at $t=5$ seconds:

At $t=5$ seconds, $F=4N$

b) When was a constant force acting on the object:

Between $t=0-3s$ and $t=7-12s$

c) Calculate the total impulse acting on the object over the 15 second interval:

The total impulse acting on the object = the area under the graph. It is convenient to divide the graph into several rectangles and triangles and then calculate the sum of these areas:

Area A = $2\times 7=14$

Area B = $\cfrac { 1 }{ 2 } \times 4\times 4=8$

Area C = $5\times 6=30$

Area D = $\cfrac { 1 }{ 2 } \times 3\times 6=9$

Total area = $14+8+30+9= 61$

Therefore, Impulse, $I=61Ns$