1© manhattan press (h.k.) ltd. 10.2 stationary waves on a stretched string

© Manhattan Press (H.K.) Ltd. 1

10.2 Stationary waves 10.2 Stationary waves on a stretched stringon a stretched string


10.2 Stationary waves on a stretched string (SB p. 131)

Stationary waves on a stretched string

Stringed instruments sound by- resonance of stretched strings




Note:

1. The nature and speed of waves formed in strings are different from those formed in air. The waves formed in strings are transverse waves, while those formed in air are longitudinal waves.

2. Stationary waves can be formed in stretched strings and closed/open pipes. Once they propagate into the air, they become progressive waves. Both the speed and wavelength are changed. Only their frequency remains unchanged.




Stationary waves of various frequencies set up in wire by plucking it at different points




(a) Fundamental frequency (plucked at midpoint)

wave reflected at P

wave reflected at Qsuperposition

of waves at midpoint




(a) Fundamental frequency (plucked at midpoint)

2 frequency lFundamenta

22

oo

oo

vvf

,

v – velocity of transverse wave along wire




(b) Frequency of first overtone (finger at midpoint, plucked at ¼ of way from one end)

o1

1 2 overtonefirst ofFrequency fvvf




(c) Frequency of second overtone (finger at 1/3 of way, plucked at 1/6 of way from one end)

o2

2 323 overtone second ofFrequency fvvf




(d) Frequency of n overtone

fn = (n + 1) fo

fo – fundamental frequency2fo, 3fo,... – second harmonic,

third harmonic, ...




(d) Frequency of n overtone

Tv )( wirestretchedon wavee transversof Speed

T – tension of wire - mass per unit length of wire

Go to

Example 1Example 1Go to

Example 2Example 2

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Example 3Example 3

Go to

Example 4Example 4

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Example 5Example 5


End


Q:Q: A wire stretched between two points 1.0 m apart is plucked near one end. What will be the three longest wavelengths shown on the vibrating wire?

Solution



Solution:Solution:

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m 0222 oo .,

m 0111 .,

m 67032

23

22 .,


Q:Q: The lowest resonant frequency for a guitar string of length 0.75 m is 400 Hz. Calculate the speed of a transverse wave on the string.

Solution



Solution:Solution:

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2o

The lowest resonant frequency is the fundamental frequency.

∴ = 0.75 =

0 = 1.50 m

Speed, v = f = 400 ×1.50 = 600 m s–1


Q:Q: Fig. (a) and (b) show two different modes of vibration for the same stretched wire under different tensions. In Fig. (a), the frequency of vibration is 20 Hz. Calculate(i) the wavelength in Fig. (a),(ii) the wave speed in Fig. (a),(iii) the wavelength in Fig. (b),(iv) the frequency in Fig. (b),(v) the mass per unit length of the wire.

Solution


Fig. (a)Fig. (b)


Solution:Solution:


12

22

2o

2

2

o

222

1o

ooo

m kg 10251040

20length unit per Mass Using(v)

Hz 8842023 2040

3202 :

12

24032

1 (b), Fig.In

12002

1 (a), Fig.In

1 Using(iv)

m 32

2301

23 (b), Fig.In (iii)

s m 0400220 (a), Fig.in speed Wave(ii)

m 022

012

(a), Fig.In (i)

..v

T,Tv

.f,/.

ff

............../

f

................

f

Tvf

,.,

..fv

.,.,

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Q:Q: The key on a piano corresponding to a note of frequency 440 Hz is pressed down very lightly so that the string is free to vibrate. When the key corresponding to a note of frequency 220 Hz is struck, it is found that the 440 Hz string emits a note of frequency 220 Hz.(a) Explain briefly the observation.(b) If the 110 Hz string is also free to vibrate, what frequencies, if any, would it emit when the 220 Hz key is struck? Solution



Solution:Solution:


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(a) When the key corresponding to the frequency of 220 Hz is struck, the 440 Hz string is forced into vibration. It vibrates with the same frequency as that which causes the vibration. Hence a note of frequency 220 Hz is emitted.

(b) If the 110 Hz string is free to vibrate, it would also be forced into vibrating with the forced frequency 220 Hz or multiples of this frequency, i.e. 440 Hz, 660 Hz, ...


Q:Q: The fundamental frequency of vibration f of a wire of length stretched by a force T is given by f = A –1 T x

where A is a constant characteristic of the wire. The following readings of f were obtained for a wire of length 0.80 m under the tensions T stated.

(a) Explain briefly one method of measuring such frequencies without using a tuning fork.(b) By a graphical method, or otherwise, find the values of A and x in the equation.


T/N 100 120 140 170 200

f/Hz 140 153 166 182 198


Q:Q: The same wire was attached at one end to a vibrator, the arm of which produced a small transverse motion. The other end passed over a pulley so that the length of wire between the vibrator arm and the pulley was 0.80 m. A mass of 16 kg was attached to the free end of the wire. It was found that for a number of frequencies of the vibrator, the wire vibrated with large amplitude at certain points along its length.(c) Explain briefly why this happened.(d) Find the three lowest frequencies at which the effect was observed, and sketch the appearance of the vibrating wire in each case.

Solution



Solution:Solution:


(a) The frequency f can be measured using a xenon stroboscope. The frequency of the xenon stroboscope is increased from zero until the vibration of the wire is frozen. When this happens, the frequency of the vibration of the wire f is equal to the frequency of the stroboscope.

(b) We have: f = A–1 Tx

Taking log, log f = log (A–1 ) + x log T .......................(1)

log (T/N) 2.000 2.079 2.146 2.230 2.301

log (f/Hz) 2.146 2.185 2.220 2.260 2.297


Solution (cont’d):Solution (cont’d):


21

21

2112

11

21

1

1

kg m 211

1000.80140 ,

equation theinto ngSubstitutiHz. 140

N, 100 when equation, in the /Hz and /Nof values theFrom

becomes

equation theHence21or 0.5

. is logagainst log ofgraph ofgradient (1),equation From

5007522502

2.1842.272graph ofgradient

/N),( logagainst /Hz)( log ofgraph theFrom

.A

,ATAf

fT

f TTAf

TAf

x

xTf

...

Tf

x


Solution (cont’d):Solution (cont’d):


(c) When the end of the wire attached to the vibrator vibrates, a transverse wave travels along the wire to the other end. At the pulley, the wave is reflected. The superposition of the incident and reflected waves produces a stationary wave on the stretched wire.The lowest frequency of the vibrator when this occurs (fo) = v/2where v = speed of the transverse wave along the wire and = length of the wire. Other frequencies when this occurs are 2fo, 3fo, ...

(d) The fundamental frequency is given by

The frequencies of the first and second overtones are 354 Hz and 531 Hz respectively.

Hz 1771016800211

o

21

1

..f

TAf

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1© manhattan press (h.k.) ltd. 10.2 stationary waves on a stretched string

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