Calculating Fractional Increase in Tension for Beat Frequency Problem

In summary: I never said it was simple. I was just explaining the concept of fractional increase and how it relates to percentages. In summary, the conversation discusses finding the fractional increase in tension of two identical piano wires to produce a frequency difference of 6.0 beats/sec. The solution involves using the equations fbeat=f2-f1, f=nv/2L, and v=\sqrt{}T/\mu to calculate the new frequency and then finding the ratio of the tensions, which is 1.02 or 102%. The concept of fractional increase is explained as a percentage increase over the original value.
  • #1
UnPhysStudent
2
0

Homework Statement



Two identical piano wires have a fundamental frequency of 600 Hz when kept under the same tension.What fractional increase in the tension of wire will lead to the occurrence of 6.0 beat/sec when both wires oscillate simultaneously?


Homework Equations



fbeat=f2-f1

f=nv/2L

v=[tex]\sqrt{}T/\mu[/tex]

The Attempt at a Solution



I used fbeat=f2-f1 to solve for f2 and ended up with 606Hz. I am really confused as to how to proceed. We were told that the answer should be 0.02 but I have know idea how to get that. Can you please give me a hint as to how to begin?

Thank you.
 
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  • #2
You can also find the ratio of the tensions (that's what it means by fractional increase)

T=µv^2=µ*(2Lf/n)^2

so then look at T2/T1.
 
  • #3
Mindscrape said:
You can also find the ratio of the tensions (that's what it means by fractional increase)

T=µv^2=µ*(2Lf/n)^2

so then look at T2/T1.

I tried doing that and at first I ended up with 1.02. After staring at it a little I released that I needed to subtract 1 from my answer to get the fractional increase.

Thanks so much for your help!
 
  • #4
Well, the ratio between them is 1.02 like you said, so T2 is 1.02 times greater than T1, or in percentages 102%. Fractional increase and percent increase is how much over 1 or 100%, respectively, the numbers are. If someone says a shirt increases in price by 250% then the shirt is really 350% of what it used to be, NewPrice=3.5*OldPrice. I personally don't see why people don't just say the shirt is 350% what it used to be, or has increased to 350% of the old price, but I'm not the one making the rules. :)
 
  • #5
if its so simple then do it yourself
 
  • #6
Luongo said:
if its so simple then do it yourself

What are you talking about?
 

1. What is a simple beat frequency problem?

A simple beat frequency problem is a mathematical problem that involves finding the difference between two frequencies. It is commonly used in physics and music to determine the interference patterns of sound waves.

2. How do you solve a simple beat frequency problem?

The solution to a simple beat frequency problem involves finding the difference between two frequencies. This can be done by subtracting the lower frequency from the higher frequency. The resulting number is the beat frequency.

3. Why are simple beat frequency problems important?

Simple beat frequency problems are important because they help us understand the relationship between different frequencies and how they interact with each other. They are also useful in various fields such as music, physics, and engineering.

4. Can simple beat frequency problems be solved using a formula?

Yes, there is a formula for solving simple beat frequency problems. It is Beat Frequency = |f1 - f2|, where f1 and f2 are the two frequencies. This formula can be used to calculate the beat frequency in hertz (Hz).

5. Are there any real-life applications of simple beat frequency problems?

Yes, simple beat frequency problems have many real-life applications. They are used in tuning musical instruments, designing audio systems, and analyzing interference patterns in sound waves. They are also used in various fields of science and technology, such as acoustics, physics, and telecommunications.

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