Physics 1C Lecture 14C "The finest words in the world are only vain sounds if you cannot understand them." --Anatole France
Standing Waves You can also create standing waves in columns of air. But in air, there are special rules regarding nodes and anti-nodes at the ends. A node must exist at a closed end of a column since the movement of particles is restricted. An anti-node must exist at an open end of a column since the particles in air have complete freedom to move. You can have three situations with columns of air: open-open, closed-closed, and open-closed.
Standing Waves The closed-closed case for air columns is exactly like the same situation as we had with strings fixed on the ends (nodes at both ends). You will use the same frequency equations: You will use the same frequency equations for the open-open case as well since you have anti-nodes at both ends. f n = n v 2L
Standing Waves But for the open-closed case, then for the longest standing wave (1 anti-node) we get: For 2 anti-nodes (second harmonic), we get:
Standing Waves For 3 anti-nodes, we get: For the nth frequency: L = n 4 λ n =1, 3, 5, 7... Recall that: λ = 4 n L v = λf So in general the frequency will be: f n = v = nv λ 4L fn = n v 4L
Beats What happens if you try to add waves that have slightly different frequencies? Sometimes their crests will line up and sometimes one crest will overlap with the other trough. The result will be constructive interference sometimes and destructive interference others.
Beats This phenomenon is known as beats. The time dependent interference causes alternations in loudness of the resulting sound wave. How often you hear the loudness is called the beat frequency (frequencies must be close). The beat frequency will merely be the subtraction of the two frequencies that are interfering with one another: f beat = f 2 f 1
Clicker Question 14C-1 An open-open tube of air supports standing waves at frequencies of 300Hz and 400Hz and no frequencies between these two. The second harmonic of this tube has a frequency of: A) 100Hz. B) 200Hz. C) 400Hz. D) 600Hz. E) 800Hz.
Clicker Question 14C-2 These two loudspeakers are in phase with each other. They emit sound waves of equal amplitude with a wavelength of 1.0 meter. At the point P indicated, the interference is: A) Completely constructive. B) Completely destructive. P C) Something in between constructive and destructive.
Wave Interference Example Two strings with linear densities of 5.0g/m are stretched over pulleys, adjusted to have vibrating lengths of 50cm, and attached to hanging blocks. The block attached to string 1 has a mass of 20kg and the block attached to string 2 has mass M. Listeners hear a beat frequency of 2.0Hz when string 1 is excited at its fundamental frequency and string 2 at its third harmonic. What is one possible value for M? Answer Here we need to make a diagram of the situation and list the known quantities.
Wave Interference Answer Making a diagram we note that the mass creates the tension to increase the wave speed: Listing the quantities that we know: f beat = 2.0Hz L 1 = L 2 = 0.50m μ 1 = μ 2 = 5.0x10-3 kg/m m 1 = 20kg
Wave Interference Answer An observer hears a beat frequency of 2Hz so: f beat = 2Hz = f 2 f 1 The frequency for the first string will be: f n = n v 2L ( ) f 1 = 1 v 1 = 2L 1 F T µ1 2L 1 f 1 = m 1 g µ1 2L 1
Wave Interference Answer The frequency for the second string will be: f 2 = 3 f n = n ( ) v 2L v 2 = 3 2L 2 f 2 = 3 Mg µ2 2L 2 F T µ2 2L 2 Put these into the beat frequency equation to find out the mass, M (remember that L 1 = L 2 and μ 1 = μ 2 ).
Wave Interference Answer f beat = 2Hz = f 2 f 1 2Hz = 3 Mg µ 2L m 1 g µ 2L 2Hz = g µ ( 3 M 20kg) 2L The factor out in front is: Eliminating the absolute value sign: g µ = 2L 9.8 N kg 5.0 10 3 kg m 2 0.5m ( ) 0.0452 kg = 3 M 4.472 kg 4.52 kg = 3 M M = 2.27kg = 44.3 1 s kg
Vibrations When I make a string move up and down, I am forcing it to vibrate. This is known as forced vibrations. Yet nearly every object will vibrate at a certain frequency which is known as the natural frequency. You can also attempt to force vibrations to an object at its natural frequency. This is known as resonance. When resonance occurs, the amplitude of vibrations will increase dramatically.
For Next Time (FNT) Start reading Chapter 21 Keep working on the homework for Chapter 14