2.  Suppose you are trying to get a swing in motion.  You've probably seen "beginners" kick their feet out at random frequencies, causing the swing to "jump" forward and backward, but not start swinging in an efficient way. Topic 4.5 Extended A – Standing waves and resonance  Although you've probably never thought of swinging in these terms, you've learned from experience that there is a natural frequency which you can apply with your feet, so that the swing can be put in the desired motion with the least effort.  Any system which oscillates will have natural frequencies at which it will oscillate most efficiently.  If an external force matches this natural frequency, the energy provided by this external force will be transferred more efficiently to the system. Less energy will be needed to "power" the oscillation.

3. Topic 4.5 Extended A – Standing waves and resonance  Consider the two waves in a plucked guitar string.  The BLUE wave is traveling leftward, and the GREEN wave is traveling rightward.  Both waves are reflected from their fixed end points (not shown), and interfere with one another.  If we look at the sum of the waves at different instances, we have the RED wave.  Note that some portions of the string, called nodes, do NOT move from their equilibrium position. N N N N N N N  Note that other portions of the string, called antinodes, undergo MAXIMUM motion. AAA A A A FYI: The guitar string actually looks like the RED wave because of the principle of superposition. FYI: We call a wave with nodes and antinodes a STANDING WAVE. The RED wave stands in place, and is the sum of two identical traveling waves with wave velocities in OPPOSITE directions. FYI: Of course, the lobes of the antinodes keep reversing positive to negative: FYI: If energy of the right frequency is added to the vibrating string, the amplitude of the antinodes will increase. This is called RESONANCE.

4.  Consider a guitar string, anchored firmly at both ends.  Since the string is fixed at its ends, the ends cannot move up and down. Topic 4.5 Extended A – Standing waves and resonance  However, the other points on the string can.  If the distance between the ends is L L  Recall that v = f, where v is the wave velocity (how fast the wave crest travels through the string), and f is the linear frequency.  Thus we have f = v v2Lv2L Fundamental Frequency of Stretched String FYI: We call the LOWEST NATURAL FREQUENCY of a vibrating system its FUNDAMENTAL FREQUENCY., we see that the wavelength of this particular vibration is 2L.

5. Topic 4.5 Extended A – Standing waves and resonance L  The next natural wavelength is illustrated, below:  The wavelength of this particular vibration is L.  The next natural wavelength is illustrated, below:  The wavelength of this particular vibration is L. 2323  If we are clever, we can deduce a pattern for the natural wavelengths of the string... 1 = L 2121 2 = L 2222 3 = L 2323 n = L 2n2n nth natural wavelength FYI: We call points that remain at the equilibrium position NODES. FYI: We call points that have maximum displacement ANTINODES. FYI: We call the set of natural wavelengths (or frequencies) the HARMONICS. First harmonic Second harmonic Third harmonic FYI: The first harmonic is also known as the FUNDAMENTAL. = Fundamental

6. Topic 4.5 Extended A – Standing waves and resonance  Since v = f, and, w e have f = v fn =fn = nv 2L The Harmonics of a Stretched String n = 1,2,3,... n = L 2n2n so that  Furthermore, it can be shown that the wave velocity of a stretched string depends on two things: the tension F in the string, and the mass per unit length  of the string. Thus v = FF The Wave Velocity of a Stretched String  In terms of the properties of the string, then, we have fn =fn = n2Ln2L FF The Harmonics of a Stretched String n = 1,2,3,...

7. Topic 4.5 Extended A – Standing waves and resonance Suppose a guitar string has a length of 1.25 meters before installation and a weight of 15 grams. (a) What is the mass per unit length  of the string? Since m = 0.015 kg, and L = 1.25 m,  = mLmL = 0.015 kg 1.25 m = 0.012 kg/m (b) If the tension is 2500 n, what is the wave velocity of the string? v = FF = 2500 0.012 = 456 m/s (c) What are the first two harmonic frequencies of this string if its ends are separated by 0.75 m? f1=f1= 1(456) 2(0.75) fn =fn = nv 2L = 304 Hz f2=f2= 2(456) 2(0.75) = 608 Hz

8. A musical octave spans a factor of two in frequency and there are twelve notes per octave. Thus notes are separated by the factor 2 1/12 or 1.059463. NotesFrequency (octaves) A55.00110.00220.00440.00880.00 A#58.27116.54233.08466.16932.32 B61.74123.48246.96493.92987.84 C65.41130.82261.64523.281046.56 C#69.30138.60277.20554.401108.80 D73.42146.84293.68587.361174.72 D#77.78155.56311.12622.241244.48 E82.41164.82329.64659.281318.56 F87.31174.62349.24698.481396.96 F#92.50185.00370.00740.001480.00 G98.00196.00392.00784.001568.00 A♭A♭ 103.83207.66415.32830.641661.28 Starting at any note the frequency to other notes may be calculated from its frequency by: Freq = note x 2 N/12, where N is the number of notes away from the starting note. N may be positive, negative or zero. For example, starting at D (146.84 Hz), the frequency to the next higher F is: 146.84 x 2 3/12 = 174.62, since F is three notes above. The frequency of A in the next lower octave is: 146.84 x 2 -17/12 = 55, since there are 17 notes from D down to the lower A. The equation will work starting at any frequency but remember that the N value for the starting frequency is zero. Fun with Frequency

9.  The following slides will show how fast a wave travels in a tight string, having a tension F and a linear mass density .  Consider a small portion of the string while the wave travels through it: Topic 4.5 Extended A – Standing waves and resonance LL  Only the string applies a force to our small mass of length  L, whose magnitude is equal to the tension F in the string: F F  Notice that at the top of the pulse, our little mass has the shape of an arc.  This leads us to suppose that if  L is sufficiently small, we can find a circle which fits it exactly:

10.  Now we observe a large- scale picture of our string, on our circle: Topic 4.5 Extended A – Standing waves and resonance LL F F  The top of the circle represents the CREST of the wave, so we know the circle is traveling at the wave speed of v to the right:  If we choose a coordinate system that is also traveling to the RIGHT at v, the picture we have drawn will not change at all, and the mass travels LEFT at v.  We know that the small mass at the top of the circle is therefore always accelerating toward the center of the circle at a centripetal acceleration of a = v2rv2r, where r is the radius of our circle. FYI: We have chosen a coordinate system that is not stationary. Is this CS an inertial CS? r r

11. Topic 4.5 Extended A – Standing waves and resonance LL F F  r r   F sin  a = v2rv2r 2F sin  = m  From Newton's 2nd we have  F = ma v2rv2r  But from the triangle above we have sin  =  L/2 r  Thus 2F = m  L/2 r v2rv2r  which reduces to F  L = mv 2

12. Topic 4.5 Extended A – Standing waves and resonance  = mLmL then m =  L.  Since  Then F =  v 2 v = FF The Wave Velocity of a Stretched String F  L = mv 2 becomes F  L =  Lv 2 FYI: If the tension increases, the velocity increases. FYI: If the mass density increases, the velocity decreases. Question: How could you show in another way that this formula is correct (at least to a constant)?