How Does Changing String Tension Affect Beat Frequency?

In summary, the fundamental frequency of a string is proportional to the square root of its tension. If the tension in one of two identical strings is increased by 2.20%, the resulting beat frequency can be found by first calculating the new tension (1.022T) and then using this value to find the new fundamental frequency (f_2). The beat frequency is then given by the difference between the two fundamental frequencies (f_2 - f_1).
  • #1
MozAngeles
101
0

Homework Statement


Two identical strings with the same tension vibrate at 630 Hz.If the tension in one of the strings is increased by 2.20% what is the resulting beat frequency?


Homework Equations



fbeat=⎮f2-f2
f=v/λ
f1=v/2L
v=√(Tension/μ)
μ=m/L

The Attempt at a Solution


I don't know where to start because with every equation i want to start off using there is something missing.
 
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  • #2
MozAngeles said:
I don't know where to start because with every equation i want to start off using there is something missing.
Start off by keeping everything in terms of variables (don't try plugging in any numbers yet). Combine your equations to determine a relationship between a string's tension and its fundamental frequency.

Once you have that, you can treat all the other variables as one big constant. And you'll be able to calculate what change in frequency corresponds to a given change in tension.
 
  • #3
so what i did was:
since f=v/λ
v=√(T/(m/L))
λ=2L
f1=√(T/(m/L))/2L
solving for Tension,
T=m4Lf12
then I plugged that into
f2=v/λ
since T2 is 2.2%, i used .022T
f2= √(.022T/μ)/2L
f2=√(.022∗m*f2*4L/(m/L))/2L
mass cancels
, then L eventually does too
and i get 93.4 for f2
then subtracting f1-f2 to get f beat i get 537, which isn't right...
So i do not know where i am going wrong? do i have the right idea going?
 
  • #4
MozAngeles said:
so what i did was:
since f=v/λ
v=√(T/(m/L))
λ=2L
f1=√(T/(m/L))/2L
solving for Tension,
T=m4Lf12
You didn't really need to solve for T. :-p But there's nothing wrong with doing so. Anyway...

The important point to gather from the above equations is that the fundamental frequency of a string is proportional to the square root of the tension. In other words,

[tex] f_1 \propto \sqrt{T} [/tex]
then I plugged that into
f2=v/λ
since T2 is 2.2%, i used .022T
Your above approach would work if the problem statement said, "If the tension in one of the strings is decreased to a value that is 2.20% of its original tension..."

But that's not what it says. It says, "If the tension in one of the strings is increased by 2.20%." That means the new tension of the string is what it was before, plus an additional 2.20%. In other words, the new tension is 1.022T.
 
  • #5
That worked.. Thank you soooo much for the help :)
 

Related to How Does Changing String Tension Affect Beat Frequency?

1. What is a wave?

A wave is a disturbance that travels through a medium, transferring energy from one point to another without actually transporting matter.

2. How are waves classified?

Waves can be classified as mechanical or electromagnetic. Mechanical waves require a medium to travel through, while electromagnetic waves can travel through a vacuum.

3. What is sound frequency?

Sound frequency refers to the number of cycles or vibrations per second that a sound wave produces. It is measured in Hertz (Hz) and determines the pitch of a sound.

4. How does sound frequency affect the human ear?

High frequency sounds (above 20,000 Hz) are typically not heard by the human ear, while low frequency sounds (below 20 Hz) are perceived as deep and bass-like. The human ear is most sensitive to sounds with frequencies between 2,000 and 5,000 Hz.

5. How is sound frequency used in technology?

Sound frequency is used in various technologies, such as ultrasound imaging, sonar, and music production. It can also be manipulated to create different sounds and effects in audio engineering.

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