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Set 56 Problem number 11


Problem

Suppose that you are holding one end of a heavy flexible cable and the other end is attached to a wall. The cable is held at some constant tension. A transverse pulse in the cable requires 1.5 seconds to travel its length.

With what frequency must you drive the end of the cable to create a wave with nodes at its two ends and only one antinode?   Assume that you are driving the end in low-amplitude SHM, which effectively makes your end a node.

Solution

To obtain the 2-node standing wave, you must be moving so as to reinforce each pulse when it returns to you. Then you will be in resonance with the standing wave: each of your movements will contribute to the energy of the wave, and the wave's amplitude will increase until it is dissipating energy (through internal friction, work against air resistance, etc.) as fast as you supply it.

When you send a 'positive' pulse down the cable (if you wish, you may regard 'positive' as meaning 'to the right'; your positive pulse would then be created by moving your end to the right; a negative pulse would of course be to the left), it remains positive until it reaches the wall, then it returns as a negative pulse. It is as if there is a 'negative' phantom pulse coming at you from the wall, with its phase precisely matched to your positive pulse so that the two pulses cancel at the wall, with the result that the wall end of the cable remains fixed (as it must).

When the negative pulse again reaches your end, it will tend to reflect in the analogous manner and travel back down the cable as a positive pulse. If you are moving your end in the positive direction at the instant the negative pulse arrives, you will reinforce its natural tendency to reflect as a positive pulse. You will be in resonance, as described above. You need not move the cable much; whatever energy you add pulse by pulse will accumulate in the wave. Your end can therefore be virtually a node. 

The time required for the pulse to travel down the cable and back is 2( 1.5 sec) = 3 sec. If you are originally at equilibrium and you begin moving your end in SHM, initially in the positive direction, then 3 seconds later the pulse will return to you. The pulse will be reinforced if you are at this instant again at the equilibrium position and moving in the positive direction. This pulse will again be reinforced on its next return if at that instant you are once more passing through equilibrium in the positive direction. If your timing is such that this precise reinforcement occurs with every return, you will remain in phase with the wave's natural behavior, and you will achieve the maximum amplitude for the SHM you are producing.

Your SHM must therefore have a period of 3 sec. The corresponding frequency is 1/( 3 sec) = .3333 Hz. 

We need to understand the relationship between the spacing of nodes and antinodes and wavelength.

Since the disturbance you send out into your original cable has exactly two antinodes in the length of the cable, the length of the cable must correspond to half of a peak-to-peak distance, or to half of a wavelength. The wavelength must therefore be twice the length of the cable.

The standing wave so produced is called the 'fundamental harmonic' of the cable. The fundamental harmonic of a standing wave can be thought of as the naturally reinforcing mode of oscillation in which the number of nodes is a minimum.

Digression (note for reference in a later problem):   Note that to maintain the wave work must be done.

To create a 3-node wave (which will have 2 antinodes)  you must send two positive pulses down the cable during the time required for the initial pulse to return.

Any motion created by timing your SHM in such a way that your cycle is opposite in phase to that of the return cycle will result in the same sort of resonance, with one antinode corresponding to each half wavelength of your cyclic pulse. Thus:

To create a 4-node wave (with 3 antinodes) you will send three positive pulses in the time required for the initial pulse to return.

To create a 5-node wave (with 4 antinodes) you will send four positive pulses in the time required for the initial pulse to return.

The frequencies are .3333, .6666, 1, and 1.333 Hz. These frequencies form the ratio sequence

It is easy to see that the frequency ratios will continue as 5/4, 6/5, ..., n/(n-1), ....

`The naturally arising frequencies of a Slinky, a chain, a string, a resonant column of air, or any other vibrating object are called 'harmonics'.

Musical instruments typically reflect certain vibrations of these strings or air columns back to the strings or air columns themselves and hence tend to reinforce those vibrations they reflect. This results in different combinations of the harmonics, and gives each instrument its characteristic sound. The shapes of the resonant cavities in your head, and the structure of your vocal cords and the tension you create in them, give your voice its distinctive character.

 

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