course Phy 202
Mr.Smith, I am working on a practice Test 2 and have a few questions-- The problem says: Traveling waves are set in a pair of long strings by a single harmonic oscillator. The waves have identical propagation velocities of 70 m/s. Both strings terminate at a short bungee cord attatched to a wall. The harmonic Oscillator is attached to the other ends of the strings in such a way that one string is 7.6 meters longer than the other. Give at least two oscillation frequencies which will produce the max motion of the bungee cord. Give at least two oscillation frequencies which will produce the min motion of the bungee cord. I have reviewed Ch. 11 and Prob Set 6 but can't figure out how to start this problem or where to go with it. Could you please point me in the right direction.
Maximum motion is achieved when the peaks of one wave arrive in conjunction with the peaks of the other.
If this happens then a whole number of peaks must fit into that extra 7.6 meters of string.
It could happen if one peak fits into the 7.6 meters, in which case the wavelength would be 7.6 meters.
Or two peaks could fit in which case the wavelength would be 1/2 * 7.6 m = 3.8 m.
From the given propagation velocity and wavelengths you can easily find the frequencies.
To produce minimum motion the peaks of one wave must arrive at the same time as the valleys of the other. This occurs when the extra 7.6 meters corresponds to half a wavelength, or to one wavelength plus half a wavelength, or in general to any whole number plus half a wavelength.
If the 7.6 m corresponds to half a wavelength then the wavelength is 2 * 7.6 m = 15.2 m.
If the 7.6 m corresponds to a whole wavelength plus half a wavelength then 1.5 wavelength, or 3/2 wavelength, corresponds to this distance and we have 3/2 * lambda = 7.6 m, so the wavelength is 2/3 * 7.6 m = 5.07 m.
Again the propagation velocity and wavelength give you the frequency.
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