A string with a fixed frequency vibrator at one end is subjected to varying tensions. When the tension is 20 N, a resonance with 3 antinodes results. What tension would cause a resonance with 2 antinodes in this string?
A) 30 N
B) 45 N
C) 80 N
D) 8.9 N
E) 13 N

Question

Quizplus · Accepted Answer

The frequency of a vibrating string is given by:
f = (1/2L) * sqrt(T/u)
where L is the length of the string, T is the tension, and u is the linear density of the string. Since the linear density is not changing, we can assume that it is constant.
If we have a resonance with 3 antinodes, then the wavelength is equal to twice the length of the string. This means that:
wavelength = 2L = 3/2 * length of string
We can rearrange the frequency equation to solve for tension:
T = (4f^2 * L^2 * u) / 1
If we plug in the values for a resonance with 3 antinodes, we get:
20 = (4f^2 * L^2 * u) / 1
If we plug in the values for a resonance with 2 antinodes, we get:
T = (4f^2 * L^2 * u) / (3/4)
We want to find the tension that causes a resonance with 2 antinodes. Since L and u are constant for this string, we can simplify this equation to:
T = (16/3) * f^2 * L^2 * u
We can solve for T by dividing this equation by the first equation:
T (2 antinodes) / T (3 antinodes) = 16/3
T (2 antinodes) = (16/3) * T (3 antinodes)
T (2 antinodes) = (16/3) * 20 = 106.7
Rounding to the nearest option gives us:
Answer: B

A String with a Fixed Frequency Vibrator at One End