Set 56 Problem number 18

Problem

Two strings are attached to the navel ring of a volunteer. The strings are held under identical 5.6 Newtons tensions and have mass density 13.2 grams/meter.

The far ends of the strings are attached to the same simple harmonic oscillator. The oscillator can be driven at 11.78 Hz, 94.29 Hz, 70.72 Hz or 35.36 Hz. If one string is 9 cm longer than the other, then at a given amplitude of oscillation, which frequency would be most likely to cause the volunteer discomfort, and which the least?

Solution

If the peaks of the two waves arrive at the ear of the volunteer simultaneously, whatever discomfort they cause will be maximized, since they will reinforce one another. If a 'peaks' of one wave arrive simultaneously with the 'valleys' of the other, the two will 'cancel out' and cause minimum discomfort.

If the difference in the distances traveled by the two waves is 1, 2, 3, ... complete wavelengths then, since the two start in phase, they will arrive in phase. If the difference in distances is 1/2, 3/2, 5/2, ... complete wavelengths then the peaks of one will arrive along with the valleys of the other.

We calculate the wavelengths corresponding to the four given frequencies:

We first calculate the wave velocity, which is v = `sqrt( T / `mu ), where T is the tension 5.6 N and `mu the mass density 13.2 grams/ meter * (1 kg / 1000 grams). A simple calculation shows that v = 424.2 m/s.

For each frequency f the wavelength is `lambda = v / f, so we obtain wavelengths

`lambda1 = 424.2 m/s / ( 11.78 cycles/sec) = 36.01 meters,

`lambda2 = 424.2 m/s / ( 94.29 cycles/sec) = 4.498 meters,

`lambda3 = 424.2 m/s / ( 70.72 cycles/sec) = 5.998 meters,

`lambda4 = 424.2 m/s / ( 35.36 cycles/sec) = 11.99 meters.

We next calculate the number of wavelengths in the distance 9 meters:

For 11.78 Hz we see that the wave on the longer string lags the other by 9 meters / ( 36.01 meters/cycle) = .2499 cycle(s).
For 94.29 Hz we see that the wave on the longer string lags the other by 9 meters / ( 4.498 meters/cycle) = 2 cycle(s).
For 70.72 Hz we see that the wave on the longer string lags the other by 9 meters / ( 5.998 meters/cycle) = 1.5 cycle(s).
For 35.36 Hz we see that the wave on the longer string lags the other by 9 meters / ( 11.99 meters/cycle) = .7506 cycle(s).

From these results it is clear which frequency gives us a whole number (e.g., 1, 2, 3, ... ) of wavelengths of path difference, which gives an integer plus a half-integer number (e.g, .5, 1.5, 2.5, ...), and which give integer plus or minute quarter-integer numbers (e.g., .25, .75, 1.25, 1.75, ...). The first will reinforce and cause maximum effect on the ear; the second will cancel and cause minimum effect; and the last will cause an intermediate effect.

Generalized Solution

If a string of mass density `mu is held at tension T, transverse waves will propagate in the string at velocity v = `sqrt( T / `mu ).

If harmonic traveling waves start out in phase along two such strings of differing length and terminate at the same point, then the wave traveling in the longer string will lag the wave traveling in the shorter string by the length difference `dL, which will also be referred to as the 'path difference'.

If the path difference is a whole number of wavelengths the waves will reinforce, and we will say that they undergo 'constructive interference'.
If the path difference is a whole number of wavelengths plus half a wavelength the waves will cancel one another, and we will say that they under 'destructive interference'.

If the wave is driven by a simple harmonic oscillator at frequency f, then the waves will have wavelength

`lambda = v / f = 1/f * `sqrt(T / `mu).

The number of wavelengths in the path difference is `dL / `lambda. We test to see if this number is a whole number or a whole number plus half, which would correspond to constructive or destructive interference respectively. The closer to these values the more nearly the behavior of the joining waves is, respectively, constructive or destructive.