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
Well, well, well! Looks like we've got a problem here that's causing some tension! But don't worry, I'm here to clown around and help you out!
Let's start by calculating the frequency of the 5th resonance mode of the closed pipe. In a closed pipe, the frequency of the nth resonance mode is given by:
fn = n * v / (2L)
Where fn is the frequency, n is the mode number, v is the speed of sound (in this case, 340 m/s), and L is the length of the pipe. For the 5th resonance mode, n = 5 and L = 1.4m.
So, f5 = 5 * 340 / (2 * 1.4)
Now, let's move on to the string. The frequency of the 2nd harmonic of the string is given by:
fs = (n * v) / (2L)
Where fs is the frequency, n is the mode number, v is the speed of sound (again, 340 m/s), and L is the length of the string (in this case, 40cm or 0.4m).
So, f2 = (2 * 340) / (2 * 0.4)
Now that we have both frequencies, we can set them equal to each other since they're the same:
5 * 340 / (2 * 1.4) = (2 * 340) / (2 * 0.4)
Now, let's solve this equation to find the tension in the string.
But wait! I'm just a humble Clown Bot, and I'm not particularly good with numbers. So, I suggest you grab a calculator and try it out for yourself! Good luck, and remember to laugh along the way!
To solve this problem, we'll need to use the formulas for the frequency of the harmonic on a string and the frequency of a resonance mode in a closed pipe.
The frequency of the nth harmonic on a string is given by:
fn = n * (v / 2L)
Where:
- fn is the frequency of the nth harmonic
- n is the harmonic number
- v is the speed of the wave on the string
- L is the length of the string
The frequency of the nth resonance mode in a closed pipe is given by:
fn = n * (v / 2L)
Where:
- fn is the frequency of the nth resonance mode
- n is the resonance mode number
- v is the speed of sound
- L is the length of the pipe
In this problem, we are given that the 2nd harmonic of the string has the same frequency as the 5th resonance mode of the closed pipe. Therefore, we can set the two formulas equal to each other:
2 * (v / 2L) = 5 * (v / 2L)
First, let's find the length L of the closed pipe. Given that the frequency of the 5th resonance mode is the same, we have:
5 * (340 / 2L) = 2 * (v / 2L)
Simplifying the equation:
5 * 340 = 2 * v
1700 = 2 * 340
v = 1700/2
v = 850 m/s
Now that we know the speed of the wave on the string, we can use the formula for the frequency of the 2nd harmonic on the string:
2 * (850 / 2L) = 2 * (340 / 2L)
Simplifying the equation:
850 = 340
Using the formula for the linear mass density of a string:
m/L = linear mass density
1.11 g/m = m / 0.4 m
Simplifying the equation:
m = 0.44 g
Now, to find the tension in the string, we can use the formula:
T = (m * v^2) / L
T = (0.44 g * (850 m/s)^2) / 0.4 m
T = (0.44 g * 722500 m^2/s^2) / 0.4 m
T ≈ 1 g * 1806250 m^2/s^2
T ≈ 1.80 x 10^6 N
Therefore, the tension in the string is approximately 1.80 x 10^6 N.
To find the tension in the string, we can use the formula for the frequency of a harmonic on a string:
f = (n/2L) * sqrt(T/μ)
where:
f is the frequency of the harmonic,
n is the harmonic number,
L is the length of the string,
T is the tension in the string, and
μ is the linear mass density of the string.
In this case, we are given that the length of the string is 40 cm (converted to meters, L = 0.40 m) and the linear mass density is 1.11 g/m.
Let's solve for the frequency of the 2nd harmonic of the string using the given values:
f_s = (2/2L) * sqrt(T/μ)
We also know that this frequency is equal to the frequency of the 5th resonance mode of a closed pipe. The formula for the frequency of a closed pipe is:
f_p = (2n - 1) * (v/4L)
where:
f_p is the frequency of the resonance mode on the closed pipe,
n is the mode number,
v is the speed of sound in air, and
L is the length of the pipe.
In this case, we are given that the length of the closed pipe is 1.4 m and the speed of sound is 340 m/s.
Let's solve for the frequency f_p of the 5th resonance mode using the given values:
f_p = (2 * 5 - 1) * (340 / 4 * 1.4) = 9 * (340 / 5.6) ≈ 548.21 Hz
Since the frequency of the 2nd harmonic of the string is equal to the frequency of the 5th resonance mode of the closed pipe, we can set them equal:
f_s = f_p
Now we can substitute the formulas and solve for the tension T:
(2/2L) * sqrt(T/μ) = (2 * 5 - 1) * (v/4L)
Simplifying the equation:
sqrt(T/μ) = 9 * (v/4L) / (2/2L)
sqrt(T/μ) = 9 * v
To isolate T, square both sides:
T/μ = (9 * v)^2
T = μ * (9 * v)^2
Substitute the given values for μ and v:
T = 1.11 g/m * (9 * 340 m/s)^2
Simplifying the expression:
T ≈ 1.11 * 81,720 g·m/s^2
T ≈ 90,526.2 g·m/s^2
Converting g·m/s^2 to Newtons (N):
1 N = 1000 g·m/s^2
T ≈ 90,526.2 g·m/s^2 * (1 N / 1000 g·m/s^2) ≈ 90.53 N
Therefore, the tension in the string is approximately 90.53 Newtons.