Putting all these proportionalities together gives: f í 1
___ l
Putting k = 1 √
___ μ or f = k
___ T
__ 2 yields the final formula:
Frequency of a stretched string f = 1
____ 2l
√
___ μ
___ T f = frequency (Hz), l = length (m), T = tension (N), μ = mass per unit length (kg m–1) QUESTIONS AND ANSWERS
3. A wire 2.5 m long and with a mass per unit length of 0.0003 kg m–1 undergoes a vibration when subjected to a tension of 9 N. Calculate the fundamental frequency of the wire.
Solution f = 1
___ 2l
√ __
__ μ ⇒ f = 1
T
________ 2(2.5)
= 0.2 √ 30 000 = 34.64 Hz _________ √
4. A 3 cm side cube of iron with a density of 8000 kg m–3 is stretched into a uniform cylindrical wire of length 2 m. Calculate the mass per unit length of this wire.
You may now complete Exercise 17B (page 202).
You may now proceed with the mandatory experiment: Investigation of the variation of fundamental frequency of a stretched string with tension (page 435).
Pipe harmonics
Different harmonics can be set up in a pipe, depending on its length and whether it is open or closed. Fig. 17.17 shows an example for a pipe closed at one end. In order to hear the harmonic, the antinode needs to be at the open end of the pipe for maximum amplitude.
When we looked at stretched strings, stationary (standing) wave patterns were drawn to illustrate the amount of movement of the
closed pipe AN pipe length = 1/4 wavelength
FIG. 17.17 In order to hear the harmonic on a pipe closed at one end, the antinode needs to be at the open end for maximum amplitude
LEAVING CERTIFICATE PHYSICS 199
_________ 0.0003
9 __________
Solution Volume = 3 cm × 3 cm × 3 cm = (3 × 10−2 m)3 = 2.7 × 10−5 m3
Mass = Volume × Density Mass = 2.7 × 10−5 × 8000 = 0.216 kg Therefore the mass per unit metre = 0.216 kg/2 = 0.108 kg m−1
NOTE You do not have to prove k = 1
_ 2 .
k is a constant of proportionality that allows us to convert a proportional relationship to a mathematical formula.
