Calculating the mass per unit length of a string based on the graph of f vs. 1/L

In summary, the task at hand is to calculate the mass per unit length of a vibrating string by analyzing a graph of frequency versus 1/L. The graph has an equation of y=9.15x + 1.03, where x represents 1/L and y represents the frequency. By comparing this to the equation f = (1/2L)sqrt(T/μ), it can be determined that the slope of the graph is equal to (1/2)sqrt(T/μ). This can be used to solve for the linear mass density of the string. However, the value of the tension (T) must be expressed in Newtons to get a meaningful result.
  • #1
buttermellow
8
0

Homework Statement



The frequency of a vibrating string is set to 15, 20, 25, 30, or 35 Hz and the length needed to attain a standing wave (mode 1) is recorded. A graph of frequency versus 1/L is recorded. Calculate the mass per unit length of the string.

The resulting graph has the equation y=9.15x + 1.03

Homework Equations



f= m/2L x (sqrt(T/mu))

m=1
T=tension= 1470 g m/s2
L=1m


The Attempt at a Solution



I assumed it had something to do with the slope, which would be equal to 1/2L x sqrt(T/mu). That doesn't make solving for mu any easier though, so what's the point? If I set 9.15 equal to this, mu comes out to be 1.09 g/m, is this right? Gah, I'm so confused!
 
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  • #2
buttermellow said:

Homework Statement



The frequency of a vibrating string is set to 15, 20, 25, 30, or 35 Hz and the length needed to attain a standing wave (mode 1) is recorded. A graph of frequency versus 1/L is recorded. Calculate the mass per unit length of the string.
Are you sure you quoted the question correctly? What does "A graph of frequency versus 1/L is recorded" mean when L = 1 m = constant? What is your independent variable?
 
  • #3
The question is from a lab, and yes, it was worded poorly.

The frequency was set and the length changed until a standing wave at the first mode was attained. After doing that for each frequency, a graph of frequency vs. 1/L was constructed (not recorded, as I said earlier). From this graph (and I assume related equations) we are supposed to find the mass per unit length of the string.
 
  • #4
For the first harmonic (fundamental) λ = 2L so that v = f(2L) = sqrt(T/μ) which gives

f = (1/2L)sqrt(T/μ) (there is no extra m multiplying the square root).

Let f = y and x = (1/L). Then

y = (1/2)sqrt(T/μ)*x

This says that if you plot y (a.k.a. f) vs. x (a.k.a. 1/L) you should get a straight line that passes through zero and has a slope equal to (1/2)sqrt(T/μ). So if you know the slope, you can find the linear mass density from

slope = (1/2)sqrt(T/μ)

Try fitting a straight line by constraining the intercept to be zero. If it doesn't work, then you will have to explain what the intercept means physically in terms of what you did in the lab.

Finally, I am not sure what you mean by "T=tension= 1470 g m/s2". To get a result that makes sense, you need to express the tension in Newtons.
 
  • #5




To calculate the mass per unit length of the string, we can use the equation y=9.15x + 1.03, where y is the frequency and x is 1/L. We can rearrange this equation to solve for m, the mass per unit length, by dividing both sides by 2L and then squaring both sides to eliminate the square root. This gives us the equation m= (y/2L)^2 x T, where T is the tension in the string.

Substituting the given values, we get m= (15/2 x 1)^2 x 1470 = 110.25 g/m. This means that for every meter of the string, it has a mass of 110.25 grams.

It is important to note that this calculation assumes a linear relationship between frequency and 1/L. If the relationship is not linear, then this method may not give an accurate result. Additionally, the tension in the string may change as the frequency is varied, which would also affect the accuracy of this calculation.
 

1. How do I calculate the mass per unit length of a string based on the graph of f vs. 1/L?

To calculate the mass per unit length of a string, you will need to plot a graph of the frequency (f) of the string's vibrations versus the inverse of its length (1/L). Then, you can use the slope of the line on the graph to determine the mass per unit length of the string, using the equation: m/L = 4Lf^2/π^2, where m is the mass per unit length, L is the length of the string, and f is the frequency.

2. What is the relationship between the graph of f vs. 1/L and the mass per unit length of a string?

The graph of f vs. 1/L shows a linear relationship, with the slope of the line representing the mass per unit length of the string. This means that as the length of the string increases, the frequency of its vibrations decreases, and the mass per unit length increases.

3. Can I use any units for length and frequency to calculate the mass per unit length of a string?

Yes, as long as the units for length and frequency are consistent, you can use any units to calculate the mass per unit length of a string. However, it is recommended to use units that are commonly used in physics, such as meters and hertz.

4. What factors can affect the accuracy of calculating the mass per unit length of a string using this method?

The accuracy of calculating the mass per unit length of a string using this method can be affected by several factors, such as experimental errors, variations in the string's properties (such as density and tension), and environmental conditions (such as temperature and humidity).

5. Can this method be used to calculate the mass per unit length of any type of string?

Yes, this method can be used to calculate the mass per unit length of any type of string, as long as it follows the linear relationship between frequency and inverse length. However, the accuracy of the calculations may vary depending on the properties of the string.

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