Beat frequency: which equation to use to find the mass of string 2

In summary, beat frequency is a phenomenon where two sound waves of slightly different frequencies interfere with each other, resulting in a pulsating pattern of sound. It is directly proportional to tension and inversely proportional to the mass of string 2. The equation used to find the mass of string 2 is m = (T/4l) * (f1^2 - f2^2), and it can be used for any string with a fixed length and tension, as long as the mass of the string is negligible. However, there may be limitations to its accuracy for strings with non-uniform density or large differences in frequency.
  • #1
Bri15
2
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Homework Statement



Two strings of equal length, L, are each clamped at both ends and are under the same
tension T. String 1 has a mass of 45.00 g and vibrates at a frequency of 512 Hz. String 2 is
slightly heavier. When the two strings are plucked at the same time, beats with a
frequency of 4.0 Hz are heard.

(i) What is the vibrational frequency of string 2?

(ii) What is the mass of string 2?

Homework Equations


Fbeat=f1-f2
f= sqrt (T/(m/L))/2L

The Attempt at a Solution


f2=508Hz (a)
(b) I do not understand which equation is applicable to find the mass
 
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  • #2
of string 2. Could you please provide more information about the setup and any other known variables?

Without any additional information, it is difficult to determine the mass of string 2. However, we can make some assumptions and use the given information to estimate the mass.

Assuming that the two strings have the same linear density (mass per unit length), we can use the equation for the frequency of a vibrating string to find the mass of string 2:

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

Where:
f = frequency of vibration
T = tension in the string
m = mass of the string
L = length of the string

We know that the tension (T) and length (L) are the same for both strings. We also know the frequency (f) of string 1 (512 Hz) and the frequency of the beats (4 Hz).

Using these values, we can set up the following equations:

512 Hz = sqrt(T/(m/L))/2L for string 1
508 Hz = sqrt(T/(m/L))/2L for string 2

Solving for T/(m/L) in both equations, we get:

T/(m/L) = (512 Hz * 2L)^2 = 524,288 Hz^2 * L^2 for string 1
T/(m/L) = (508 Hz * 2L)^2 = 515,584 Hz^2 * L^2 for string 2

Since we know that the tension (T) and length (L) are the same for both strings, we can set these two equations equal to each other and solve for the mass of string 2:

524,288 Hz^2 * L^2 = 515,584 Hz^2 * L^2

Dividing both sides by L^2, we get:

524,288 Hz^2 = 515,584 Hz^2

Taking the square root of both sides, we get:

724 Hz = 718 Hz

Since this is not possible, we can conclude that our initial assumption (that the two strings have the same linear density) is not correct. This means that we cannot accurately determine the mass of string 2 without more information about the setup.
 

1. What is beat frequency?

Beat frequency is a phenomenon where two sound waves of slightly different frequencies interfere with each other, resulting in a pulsating pattern of sound with a frequency equal to the difference between the two original frequencies.

2. How is beat frequency related to the mass of string 2?

The beat frequency is directly proportional to the tension and inversely proportional to the mass of string 2. This means that as the mass of string 2 increases, the beat frequency decreases.

3. What equation can be used to find the mass of string 2?

The equation used to find the mass of string 2 in a beat frequency experiment is: m = (T/4l) * (f1^2 - f2^2), where m is the mass of string 2, T is the tension in the string, l is the length of the string, and f1 and f2 are the two frequencies used.

4. Can this equation be used for any type of string?

The equation can be used for any string with a fixed length and tension. However, the mass of the string should be small enough to be negligible compared to the mass of string 2.

5. Are there any limitations to using this equation?

This equation assumes that the string has a uniform linear density and that the frequencies used are relatively close to each other. It may not be accurate for strings with non-uniform density or for large differences in frequency.

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