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Waves_03 1 Two sine waves travelling in opposite directions  standing wave Some animations courtesy of Dr. Dan Russell, Kettering University TRANSVERSE.

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Presentation on theme: "Waves_03 1 Two sine waves travelling in opposite directions  standing wave Some animations courtesy of Dr. Dan Russell, Kettering University TRANSVERSE."— Presentation transcript:

1 waves_03 1 Two sine waves travelling in opposite directions  standing wave Some animations courtesy of Dr. Dan Russell, Kettering University TRANSVERSE WAVES ON STRINGS

2 Travelling transverse waves, speed of propagation, wave function, string tension, linear density, reflection (fixed and free ends), interference, boundary conditions, standing waves, stationary waves, SHM, string musical instruments, amplitude, nodes, antinodes, period, frequency, wavelength, propagation constant (wave number), angular frequency, normal modes of vibrations, natural frequencies of vibration, fundamental, harmonics, overtones, harmonic series, frequency spectrum, radian, phase, sinusoidal functions waves_03: MINDMAP SUMMARY - TRANSVERSE WAVES ON STRINGS 2

3 3 The string (linear density  ) must be under tension F T for wave to propagate increases with increasing tension F T decreases with increasing mass per unit length  independent of amplitude or frequency TRANSVERSE WAVES ON STRINGS Wave speed v (speed of propagation) linear density

4 4 Problem 1 A string has a mass per unit length of 2.50 g.m -1 and is put under a tension of 25.0 N as it is stretched taut along the x-axis. The free end is attached to a tuning fork that vibrates at 50.0 Hz, setting up a transverse wave on the string having an amplitude of 5.00 mm. Determine the speed, angular frequency, period, and wavelength of the disturbance. [Ans: 100 m.s -1, 3.14x10 2 rad.s -1, 2.00x10 -2 s, 2.00 m] use the ISEE method

5 5 Solution 1 F = 25.0 N  = 2.50 g.m -1 = 2.50×10 -3 kg.m -1 f = 50.0 Hz A = 5.00 mm = 5×10 -3 m v = ? m.s -1  = ? rad.s -1 T = ? s = ? m Speed of a transverse wave on a string speed of a wave

6 6 Pulse on a rope When pulse reaches the attachment point at the wall the pulse is reflected If attachment is fixed the pulse inverts on reflection If attachment point can slide freely of a rod, the pulse reflects without inversion If wave encounters a discontinuity, there will be some reflection and some transmission Example: two joined strings, different . What changes across the discontinuity - frequency, wavelength, wave speed?

7 Reflection of waves at a fixed end Reflected wave is inverted  PHASE CHANGE Reflection of waves at a free end Reflected wave is not inverted ZERO PHASE CHANGE 7

8 8 Refection of a pulse - string with boundary condition at junction like a fixed end CP 510

9 9 Refection of a pulse - string with boundary condition at junction like a free end CP 510

10 10 Refection of a pulse - string with boundary condition at junction like a fixed end Refection of a pulse - string with boundary condition at junction like a free end

11 11 STANDING WAVES If we try to produce a traveling harmonic wave on a rope, repeated reflections from the end produces a wave traveling in the opposite direction - with subsequent reflections we have waves travelling in both directions The result is the superposition (sum) of two waves traveling in opposite directions The superposition of two waves of the same amplitude travelling in opposite directions is called a standing wave Examples: transverse standing waves on a string with both ends fixed (e.g. stringed musical instruments); longitudinal standing waves in an air column (e.g. organ pipes and wind instruments)

12 12 oscillation Standing waves on strings CP 511 each point oscillates with SHM, period T = 2  /  Two waves travelling in opposite directions with equal displacement amplitudes and with identical periods and wavelengths interfere with each other to give a standing (stationary) wave (not a travelling wave - positions of nodes and antinodes are fixed with time) amplitude

13 13 A string is fixed at one end and driven by a small amplitude sinusoidal driving force f d at the other end. The natural frequencies of vibration of the string (nodes at each end) are f o = 150 Hz, 300 Hz, 450 Hz, 600 Hz,... The string vibrates at the frequency of the driving force. When the string is excited at one of its natural frequencies, large amplitude standing waves are set up on the string (resonance). f d = 150 Hzf d = 200 Hzf d = 450 Hz fundamental3 rd harmonic

14 14 String fixed at both ends A steady pattern of vibration will result if the length corresponds to an integer number of half wavelengths In this case the wave reflected at an end will be exactly in phase with the incoming wave This situations occurs for a discrete set of frequencies CP 511 Speed transverse wave along string Natural frequencies of vibration Boundary conditions  Standing waves on strings

15 15 Why do musicians have to tune their string instruments before a concert? CP 518

16 16 node antinode Fundamental CP 518 Modes of vibrations of a vibrating string fixed at both ends Natural frequencies of vibration

17 17 N = 1 fundamental or first harmonic 1 = 2L f 1 = (1/2L).  (F T /  ) Harmonic series N th harmonic or (N-1) th overtone N = 2L / N = 1 / N f N = N f 1 N = 2 2 nd harmonic (1st overtone) 2 = L = 1 / 2 f 2 = 2 f 1 N = 3 3 nd harmonic (2 nd overtone) 3 = L = 3 / 2 f 3 = 3 f 1 CP Resonance (“large” amplitude oscillations) occurs when the string is excited or driven at one of its natural frequencies.

18 18 violin – spectrum viola – spectrum CP 518

19 Problem solving strategy: I S E E I dentity: What is the question asking (target variables) ? What type of problem, relevant concepts, approach ? S et up: Diagrams Equations Data (units) Physical principals E xecute: Answer question Rearrange equations then substitute numbers E valuate: Check your answer – look at limiting cases sensible ? units ? significant figures ? PRACTICE ONLY MAKES PERMANENT 19

20 20 Problem 2 A guitar string is 900 mm long and has a mass of 3.6 g. The distance from the bridge to the support post is 600 mm and the string is under a tension of 520 N. 1 Sketch the shape of the wave for the fundamental mode of vibration 2 Calculate the frequency of the fundamental. 3 Sketch the shape of the string for the sixth harmonic and calculate its frequency. 4 Sketch the shape of the string for the third overtone and calculate its frequency.

21 Solution 2 L 1 = 900 mm = m m = 3.6 g = 3.6  kg L = 600 mm = m F T = 520 N  = m / L 1 = (3.6  / 0.9) kg.m -1 = kg.m -1 v =  (F T /  ) =  (520 / 0.004) m.s -1 = m.s -1 1 = 2L = (2)(0.600) m = m Fundamental frequency f 1 = v / 1 = (360.6 / 1.2) Hz = 300 Hz f N = N f 1 sixth harmonic N = 6 f 6 = (6)(300) Hz = 1800 Hz = 1.8 kHz third overtone = 4th harmonic N = 4 f 4 = (4)(300) Hz = 1200 Hz = 1.2 kHz

22 22 Problem 3 A particular violin string plays at a frequency of 440 Hz. If the tension is increased by 8.0%, what is the new frequency?

23 Solution 3 f A = 440 Hz f B = ? Hz F TB = 1.08 F TA A = B  A =  B L A = L B N A = N B v = f v =  (F T /  ) string fixed at both ends L = N /2 = 2L / N natural frequencies f N = N v / 2L = (N / 2L).  (F T /  ) f B / f A =  (F TB / F TA ) f B = (440)  (1.08) Hz = 457 Hz

24 24 Why does a tree howl? The branches of trees vibrate because of the wind. The vibrations produce the howling sound. N A Fundamental mode of vibration Problem 4 Length of limb L = 2.0 m Transverse wave speed in wood v = 4.0  10 3 m.s -1 Fundamental L = / 4 = 4 L v = f f = v / = (4.0  10 3 ) / {(4)(2)} Hz f = 500 Hz

25 Standing waves in membranes NB the positions of the nodes and antinodes

26 26 CHLADNI PLATES

27 Some of the animations are from the web site


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