waves_03

TRANSVERSE WAVES ON STRINGS

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

 waves_03: MINDMAP SUMMARY - TRANSVERSE WAVES ON STRINGS 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

v



F T

  

m s L s v



f

 

T k

  2 

f

 2 

T T

 1

f k

 2  2 



TRANSVERSE WAVES ON STRINGS Wave speed

v

(speed of propagation)

The string (linear density  ) must be under tension

F T

for wave to propagate

v



F T

 linear density  

m s L s

• increases with increasing tension

F T

• decreases with increasing mass per unit length  • independent of amplitude or frequency 3

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 the disturbance.

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 use the ISEE method [Ans: 100 m.s

-1 , 3.14x10

2 rad.s

-1 , 2.00x10

-2 s, 2.00 m] 4

Solution 1

F A

= 25.0 N  = 2.50 g.m

-1 = 5.00 mm = 5×10 -3 m

v

= 2.50×10 -3 = ? m.s

-1  kg.m

-1

f

= 50.0 Hz = ? rad.s

-1

T

= ? s  = ? m Speed of a transverse wave on a string speed of a wave

v



T



v



f



v



T

   2  

f

 25  3 m.s

-1  100 m.s

-1 -1 

T

 1

f

 1 50 s   2 s

v



f

  

f v

 100 50 m  2.00 m  2 -1 5

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

?

6

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

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

Incident pulse Reflected pulse Reflected wave  rad (180°) out of phase with incident wave Transmitted pulse Heavy string exerts a downward force on light string when pulse arrives

8 CP 510

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

Incident pulse Reflected pulse Transmitted pulse Reflected wave: in phase with incident wave, 0 rad or 0°C phase difference Heavy string pulls light string up when pulse arrives, string stretches then recovers producing reflected pulse 9 CP 510

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 10

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

) 11

Standing waves on strings

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)  

A

sin(

kx

 

t

)



A

sin(

kx kx

) cos(



t

)

 

t

)

amplitude oscillation

each point oscillates with SHM, period

T

= 2  /  CP 511 12

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 Hz fundamental

f d

= 450 Hz 3 rd harmonic

f d

= 200 Hz 13

Standing waves on strings

• 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 Boundary conditions 

L



N

 2   2

N L v

 

N v

2

L

Speed transverse wave along string

v



F

T  Natural frequencies of vibration

f N



v



N

 1 2

L F

T  CP 511 14

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

different string  Finger board bridges - change

L

Body of instrument (belly) resonant chamber - amplifier tuning knobs (pegs) - adjust

F

T  

L f

1  1 2

L F

T 

v



F

T 

f N



N f

1

f

 1

L F

T 

N

 1, 2, 3,...

CP 518 15

Modes of vibrations of a vibrating string fixed at both ends Natural frequencies of vibration



f

 2

L



N v

 

N v

2

L

 Fundamental

f

1 

v

2

L



L



N

 2

f N



v



N

 1 2

L f N



Nf

1

F

T  16

node



antinode



CP 518

Harmonic series

120

N

th harmonic or (

N

-1) th overtone 

N

= 2

L

/

N

=  1 /

N f N

=

N f

1 100 80

N

= 3 3 nd harmonic (2 nd overtone)  3 =

L

=  3 / 2

f

3 = 3

f

1 60 40

N

= 2 2 nd harmonic (1st overtone)  2 =

L

=  1 / 2

f

2 = 2

f

1 20

N

= 1

fundamental

 1 = 2

L

f or

first harmonic

1 = (1/2

L

).

 (

F

T /  ) 0

0 0.1

0.2

0.3

0.4

0.5

0.6

Resonance position along string

(“large” amplitude oscillations) occurs when the string is excited or driven at one of its natural frequencies.

0.7

0.8

0.9

1

17 CP 518 22 23

1 0 3 2 10 9 8 5 4 7 6

1 2 3 4

violin – spectrum 50 -10 -20 -30 -40 40 30 20 10 0 -50

0 5 6 7 8 9 10 11 12 13 14 15 harmonics (fundamental f 1 = 440 Hz) 0.002

viola – spectrum 2 0 6 4 10 8 14 12

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 harmonics (fundamental f 1 = 440 Hz)

50 40 -10 -20 -30 -40 30 20 10 0 -50

0 0.002

0.004

time t (s) 0.006

0.004

time t (s) 0.006

0.008

0.008

CP 518 18

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

PRACTICE ONLY MAKES PERMANENT

E

xecute: Answer question Rearrange equations then substitute numbers

E

valuate: Check your answer – look at limiting cases sensible ?

units ?

significant figures ?

19

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.

20

Solution 2

L

1 = 900 mm = 0.900 m

m

= 3.6 g = 3.6

 10 -3 kg

L



v

 1 = 600 mm = 0.600 m

F

T =

m

=  (

F

T /

L

1 /  = (3.6

 10 -3 ) =  = 520 N / 0.9) kg.m

(520 / 0.004) m.s

-1 -1 = 0.004 kg.m

-1 = 360.6 m.s

-1 = 2

L

= (2)(0.600) m = 1.200 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 f

4 = 4 = (4)(300) Hz = 1200 Hz =

1.2 kHz

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?

22

Solution 3

f

A = 440 Hz

f

B = ? Hz

v F

TB  A = 1.08

F

TA =  B  A =  B =

f



v

=  (

F

T /  )

L

A =

L

B

N

A =

N

B string fixed at both ends

L

=

N

 /2  = 2

L

/

N

natural frequencies

f N

=

N v

/ 2

L

= (

N

/ 2

L

).

 (

F

T /  )

f

B /

f

A =  (

F

TB /

F

TA )

f

B = (440)  (1.08) Hz =

457 Hz

Problem 4

Why does a tree howl?

The branches of trees vibrate because of the wind.

The vibrations produce the howling sound.

N A Length of limb

L

= 2.0 m Transverse wave speed in wood

v f = v

= 4.0

 10 3 Fundamental

L v = f

 /  m.s

=  -1 / 4  = 4

L

= (4.0  10 3 ) / {(4)(2)} Hz

f

= 500 Hz Fundamental mode of vibration 24

Standing waves in membranes

NB the positions of the nodes and antinodes http://www.isvr.soton.ac.uk/spcg/tutorial/tutorial/Tutorial_files/Web-standing-membrane.htm

CHLADNI PLATES

26

Some of the animations are from the web site http://paws.kettering.edu/~drussell/demos.html