A string is attached to a wall and vibrates back and forth, as in Figure 17.18. The vibration frequency and length of the string are fixed. The tension in the string is changed, and it is observed that at certain values of the tension a standing wave pattern develops. Account for the fact that no standing waves are observed once the tension is increased beyond a certain value.
The time required to create a new wave cycle does not equal the time taken by a cycle to travel the entire length of the string.
Repeated reinforcement between newly created cycles causes a zero amplitude standing wave.
The frequency of vibrating cycles is sufficiently high that they all cancel each other.
When the tension is increased beyond the value for which = 2L, the string cannot contain an integer number of half wavelengths.
The statement "the frequency of vibrating cycles is sufficently high.." is meaninless. The frequency is fixed, as is the length.
The fundamental resonant frequence is a function of Tension, and Length. As tension goes up, so does the fundamental resonant freq. Of course, there are harmonics which ,i> could resonate.
Now letting F be the fixed frequency, and f the funamental resonant frequency, what happens when
f<F? Standing waves occur when n*f=F
But when f>F, then n*f > F, so there can be no resonance: The stimulation of F is below the lowest possible frequency, as well as below the possible harmonic frequencies.
The string could still resonate, but not with the fixed vibration frequcny. Plucking could do it, as well as stimulating it with square or triangular waves.
If
So you're saying that:
The time required to create a new wave cycle does not equal the time taken by a cycle to travel the entire length of the string.
is a true statement according to the question. I am kind of confused. I had ruled out the second and third option but I was torn between the first and fourth.
yes, true. The time taken for the wave to travel the length and back is too slow to resonate with the exciting frequency on the string.