Frequency of a Standing Wave on a String

In summary, the conversation discusses the relationship between the frequency and tension of a standing wave on a string. It is shown that when tension is increased by a small amount, the frequency changes by an amount deltaf, such that deltaf/f = .5 * deltaT/T. This is due to the fact that both the wavelength and frequency are functions of tension, as shown by the equation v=sqrt(T/Mu) and f= (1/lambda)*sqrt(T/Mu). This is further discussed in regards to the wavelength remaining the same when tension is increased, as both lambda and f are functions of T.
  • #1
bcjochim07
374
0

Homework Statement


The frequency of a standing wave on a string is f when the string's tnesion is T. If the tension is changed by the small amount deltaT, witout changing the length, show tat the frequency changes by an amount deltaf, such that

deltaf/f = .5 * deltaT/T


Homework Equations





The Attempt at a Solution



v=sqrt(T/Mu)

f= (1/lambda)*sqrt(T/Mu) When tension is increased, the wavelength will still be the same

f+deltaf=(1/lambda)*sqrt((T+deltaT)/Mu)
so delta f=(1/lambda)*sqrt((T+deltaT)/Mu)-f

deltaf/f = [(1/lambda)*sqrt((T+deltaT)/Mu)-f]/((1/lambda)*sqrt(T/Mu))

deltaf/f = sqrt(T+deltaT)/sqrt(T) -1

But I can't get it simplified any more than this
 
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  • #2
any thoughts on this one?
 
  • #3
bcjochim07 said:
v=sqrt(T/Mu)

f= (1/lambda)*sqrt(T/Mu) When tension is increased, the wavelength will still be the same

If [itex]\lambda = \frac{v}{f}[/itex] and [itex] \frac{dv}{dT}=\frac{1}{2\sqrt{T\mu}}[/itex], then how is it that the wavelength will be the same when tension is increased?

Regards,

Bill
 
  • #4
It seems to me that if both f and v increase by some factor, that wavelength should remain the same, and if there is one wave still on the string when the tension changes very slightly, how could the wavelength change?
 
  • #5
bcjochim07 said:
how could the wavelength change?

Because [itex]\frac{d\lambda}{dT}\neq 0[/itex]. Therefore, both [itex]\lambda[/itex] and f are functions of T.

Regards,

Bill
 

1. What is the frequency of a standing wave on a string?

The frequency of a standing wave on a string is determined by the speed of the wave and the length of the string. It can be calculated using the equation f = n(v/2L), where f is the frequency, n is the harmonic number, v is the speed of the wave, and L is the length of the string.

2. How does the length of the string affect the frequency of a standing wave?

The length of the string directly affects the frequency of a standing wave. As the length of the string increases, the frequency decreases. This is because the longer the string, the longer it takes for the wave to travel along its length, resulting in a lower frequency.

3. What is the relationship between the harmonic number and the frequency of a standing wave?

The harmonic number, or the number of nodes in the standing wave, is directly related to the frequency. As the harmonic number increases, the frequency also increases. This means that the higher the harmonic number, the more nodes there are in the standing wave and the higher the pitch of the sound produced.

4. Can the frequency of a standing wave on a string be changed?

Yes, the frequency of a standing wave on a string can be changed by altering either the length of the string or the tension applied to the string. Changing the length of the string will directly affect the frequency, while changing the tension will indirectly affect the frequency by changing the speed of the wave.

5. How is the frequency of a standing wave on a string measured?

The frequency of a standing wave on a string can be measured using a frequency meter or a tuner. These devices can detect and display the frequency of a sound wave, allowing for accurate measurement of the standing wave's frequency. Alternatively, the frequency can also be calculated using the equation f = n(v/2L) if the necessary values are known.

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