Refraction, Diffraction and Resonance - Learn

Reflection, Diffraction and Resonance

Reflection:

A common phenomenon of soundwaves is the reflection of a soundwave off some surface creating what we know as echoes. Echoes are fascinating for people opposite cliffs and mountains who might yell out and wait to hear the echo. They are used by bats to navigate, porpoises to find schools of fish. Scientific research and exploration vessels also use echoes to map ocean depths or find rock formations which may contain oil or natural gas. Commonly used forms of echoes in medicine include X-rays and ultrasounds.

Diffraction:

The diffraction of sound waves is what allows people to hear music or people talking in other rooms. Long wavelength (low frequency) sounds diffract more than short (high frequency) wavelength sounds. This allows some animals in nature to communicate over much greater distances than others. For example, owls are able to communicate across long distances because their low frequency and therefore long wavelength hoots are able to diffract around forest trees and carry farther than the short wavelength tweets of songbirds. Elephants emit infrasonic waves (below human hearing threshold) of very low frequency to communicate over long distances to each other – up to 30 kilometres as the sound transmits through the ground. Bats, being blind, use high frequency (low wavelength) ultrasonic waves (frequency of about 50 000 Hz) in order to navigate and to hunt their prey. As the wavelength of a wave becomes smaller than the obstacle that it encounters, the wave is no longer able to diffract around the obstacle, but instead, it will reflect off the obstacle. The smallest object they will be able to detect will be about the same size as the wavelength of the sound they emit.

Resonance:

Acoustic resonance refers to the amplification of a sound using a system whose frequency matches one of its own natural frequencies of vibration – also call a resonance frequency. Acoustic resonance is an important consideration for instrument builders. Shapes of pianos, acoustic guitars and wind instruments are designed in such a way as to maximise acoustic resonance.

Masses Connected by Vertical Strings

Masses connected by vertical strings is another problem that requires both Newton’s 2nd and 3rd laws. Consider two objects A and B, with masses of 2kg and 5kg respectively as shown below.

Let’s examine two situations:

a) Objects A and B are stationary

b) Objects are accelerating up at $2\cfrac { m }{ { s }^{ 2 } }$

a) If the objects are stationary, the net force on each object = 0. The forces involved will be the tension in the string. The tension in each string is equal to the weight that they support:

$T_{ 1 }$ = the weight of object A:

$W=mg$

$W=2\times 9.8$

$W=19.6\quad N$

$T_{ 2 }$ = the weight of object A + object B:

$W=mg$ . This is due to the string at $T_{ 2 }$ supporting object A and B.

$W=7\times 9.8$

$W=68.6\quad N$

Therefore; $T_{ 1 }=19.6N$ and $T_{ 2 }=68.6N$

b) If the objects are accelerating up at $2\cfrac { m }{ { s }^{ 2 } }$ , we can determine the net force acting on each object using $F=ma$ . We can also determine the tension in each string using: $T=W-ma$ . This equation involves vectors and down is best noted as positive. The acceleration will be negative if directed upwards or negative if directed downwards. Note that a positive or negative result for tension is not important because tension is a force that acts in both directions on a string. Essentially this means that the equation: $T=W-ma$ gives us a magnitude for the tension.

If we consider the acceleration of each object we can determine the net force, $F_{ net }$ acting on each object:

$F_{ net }=ma$

$F_{ net }(A)=2\times 2=4N$

$F_{ net }(B)=5\times 2=10N$

Observing the diagram below, we can see that: (we will note down as positive)

$T_{ 1 }=W-ma$

$T_{ 1 }=(2\times 9.8)-(2\times -2)$

$T_{ 1 }=(19.6)-(-4)$

$T_{ 1 }=23.6N\quad$

$T_{ 2 }=W-ma$

$T_{ 2 }=((2+5)\times 9.8)-((2+5)\times -2)$

$T_{ 2 }=(7\times 9.8)-(7\times -2)$

$T_{ 2 }=(68.6)-(-14)$

$T_{ 2 }=82.6N$

Example 1:

Two objects of masses 3kg and 10kg are connected by vertical strings as shown below. The objects are stationary. Calculate:

a) The net force acting on each mass

b) The tension in each string

a) If the objects are stationary, the net force on each object = 0

b) The tension in each string is equal to the weight that they support:

$T_{ 1 }$ = the weight of the 3kg mass:

$W=mg$

$W=3\times 9.8$

$W=29.4\quad N$

$T_{ 2 }$ = the weight of both masses:

$W=mg$ .

$W=13\times 9.8$

$W=127.4\quad N$

Therefore; $T_{ 1 }=29.4N$ and $T_{ 2 }=127.4N$

Example 2:

Two objects of masses 5kg and 12kg are connected by vertical cables as shown below. The objects are accelerating upwards at $1.5\cfrac { m }{ { s }^{ 2 } }$ . Calculate:

a) The tension in each string:

we will note down as positive:

$T_{ 1 }=W-ma$

$T_{ 1 }=(5\times 9.8)-(5\times -1.5)$

$T_{ 1 }=(49)-(-7.5)$

$T_{ 1 }=56.5N\quad$

$T_{ 2 }=W-ma$

$T_{ 2 }=((5+12)\times 9.8)-((5+12)\times -1.5)$

$T_{ 2 }=(17\times 9.8)-(17\times -1.5)$

$T_{ 2 }=(166.6)-(-25.5)$

$T_{ 2 }=191.5N$

Example 3:

Two objects of masses 2kg and 4.5kg are connected by vertical cables as shown below. The objects are accelerating downwards at $3\cfrac { m }{ { s }^{ 2 } }$ . Calculate: