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.
a) The frequency of vibrating cycles is sufficiently high that they all cancel each other.
b)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.
c) When the tension is increased beyond the value for which = 2L, the string cannot contain an integer number of half wavelengths.
d)Repeated reinforcement between newly created cycles causes a zero amplitude standing wave.
I think when the frquency is sufficiently high they'll cancel out, because frequency if proportional to tension.
Please check my answer
Your answer is wrong.
http://hyperphysics.phy-astr.gsu.edu/hbase/waves/string.html
What happens if Tension gets greater than the FIXED vibration frequency?
the string will obviosly break DUH!!!
so i think the answer will be d.
I apologize, but your answer is incorrect. The correct answer is c) When the tension is increased beyond the value for which λ = 2L, the string cannot contain an integer number of half wavelengths.
This phenomenon is known as the principle of resonance. In a standing wave pattern, the length of the string must be an integer multiple of half the wavelength of the wave. This means that the tension in the string must be adjusted in such a way that the wavelength fits perfectly within the length of the string.
When the tension is increased beyond the value for which λ = 2L, the wavelength becomes shorter and cannot fit within the length of the string. As a result, no standing waves can form.
To understand why this happens, let's look at the relationship between tension, frequency, and wavelength. The speed of a wave traveling through a string is given by v = √(T/μ), where T is the tension in the string, and μ is the linear mass density of the string. The frequency of the standing wave is determined by the number of wavelengths that fit within the length of the string. This can be expressed as f = v/λ. Rearranging these equations, we get λ = v/f, and substituting v = √(T/μ), we get λ = √(T/μ) / f.
Now, a standing wave pattern is formed when the length of the string, L, is equal to an integer multiple of half the wavelength, λ/2. Mathematically, this can be expressed as L = n(λ/2), where n is an integer. Substituting the expression for wavelength, we get L = n[(√(T/μ) / f)/2], which simplifies to 2L = (n/f)√(T/μ). Since n is an integer, the condition for a standing wave pattern to form is 2L = (n/f)√(T/μ).
From this equation, we can see that as the tension, T, increases, the value of (n/f)√(T/μ) also increases. Once the tension exceeds the value for which λ = 2L, the string cannot contain an integer number of half wavelengths, and no standing waves can form.
Therefore, the correct answer is c) When the tension is increased beyond the value for which λ = 2L, the string cannot contain an integer number of half wavelengths.