The 2nd harmonic of a string of length 40cm and linear mass density 1.11g.m has the same frequency as the 5th resonance mode of a closed pipe of length 1.4m. Find the tension in the string.
use 340 for speed of sound
To find the tension in the string, we need to use the formula for the frequency of standing waves on a string, as well as the formula for the frequency of resonance modes in a closed pipe.
Let's start by finding the frequency of the 2nd harmonic on the string.
The frequency of the nth harmonic on a string is given by:
fn = (n * v) / (2L)
where fn is the frequency, n is the harmonic number, v is the speed of the wave, and L is the length of the string.
We are given that the length of the string is 40cm, which is equal to 0.4m. We are also given the speed of the wave as 340m/s.
Substituting these values into the formula, we can find the frequency of the 2nd harmonic on the string:
f2 = (2 * 340) / (2 * 0.4) = 850 Hz
Now, let's find the frequency of the 5th resonance mode in the closed pipe.
The frequency of the nth resonance mode in a closed pipe is given by:
fn = (n * v) / (4L)
where fn is the frequency, n is the resonance mode number, v is the speed of sound, and L is the length of the pipe.
We are given that the length of the pipe is 1.4m and the speed of sound is 340m/s.
Substituting these values into the formula, we can find the frequency of the 5th resonance mode:
f5 = (5 * 340) / (4 * 1.4) = 607.14 Hz
Since we are given that the frequency of the 2nd harmonic of the string is equal to the frequency of the 5th resonance mode of the pipe, we can set these two frequencies equal to each other:
850 Hz = 607.14 Hz
Now, let's solve for the tension in the string.
The frequency of a standing wave on a string is given by:
fn = (1 / 2L) * sqrt(T / μ)
where fn is the frequency, L is the length of the string, T is the tension, and μ is the linear mass density of the string.
We are given that the linear mass density of the string is 1.11 g/m, which is equal to 0.00111 kg/m. We need to convert it to kg/m.
Substituting these values into the formula, we can solve for the tension:
(1 / 2 * 0.4) * sqrt(T / 0.00111) = 850 Hz
Simplifying this equation, we get:
sqrt(T / 0.00111) = (2 * 0.4 * 850)
T / 0.00111 = (2 * 0.4 * 850)^2
T = (2 * 0.4 * 850)^2 * 0.00111
Calculating this equation, we find:
T ≈ 231.98 N
Therefore, the tension in the string is approximately 231.98 N.