Find lowest resonant frequence on this string

In summary, a string with fixed supports separated by 75 cm has resonant frequencies of 420 and 315 Hz with no intermediate frequencies. To find the fundamental frequency, we can use the equations f_n=nf_1 and f_{n+1}=(n+1)f_1 and solve for v using the equation f_n=\frac{nv}{2L}. Then, we can plug the value of v into the first harmonic equation to find the wave speed.
  • #1
Saladsamurai
3,020
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A string that is stretched between fixed supports separated by 75 cm has resonant frequencies of 420 and 315 Hz, with no resonant intermediate frequencies. What are (a) the lowest resonant frequency and (b) the wave speed?

Okay: So I know that since there are no intermediate resonant frequencies, then the two givens are of nth harmonic and nth+1 harmonic.

I also know that [tex]f_n=nf_1[/tex] and [tex]f_{n+1}=(n+1)f_1[/tex]

so I need to find the fundamental frequency right? I am assuming this will involve 2 equations in 2 unknowns...
 
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  • #2
Should I use [tex]f_n=\frac{nv}{2L}[/tex] in a system of equations to solve for v and then plug that into find the 1st harmonic?...how bout I try it instead of asking?!
 
  • #3


First, let's define some variables. Let L be the length of the string (75 cm), n be the harmonic number, and f_1 be the fundamental frequency. We also know that the first resonant frequency (f_n) is 420 Hz and the second resonant frequency (f_{n+1}) is 315 Hz.

(a) To find the lowest resonant frequency, we need to find the value of n. We can set up the following equation:

f_n=nf_1

Substituting in the given values, we get:

420 Hz = n(f_1)

Similarly, for the second resonant frequency, we have:

315 Hz = (n+1)(f_1)

Now we have two equations and two unknowns (n and f_1). We can solve for n by dividing the second equation by the first:

315 Hz / 420 Hz = (n+1)(f_1) / n(f_1)

0.75 = (n+1)/n

0.75n = n+1

0.25n = 1

n = 4

Now that we know the value of n, we can plug it back into either of the original equations to solve for f_1:

420 Hz = 4(f_1)

f_1 = 105 Hz

Therefore, the lowest resonant frequency is 105 Hz.

(b) To find the wave speed, we can use the equation v = λf, where v is the wave speed, λ is the wavelength, and f is the frequency. We know that the wavelength is twice the length of the string (75 cm), so λ = 2(0.75 m) = 1.5 m. Plugging in the values for f (105 Hz), we get:

v = (1.5 m)(105 Hz)

v = 157.5 m/s

Therefore, the wave speed is 157.5 m/s.
 

What is the lowest resonant frequency on a string?

The lowest resonant frequency on a string is the fundamental frequency, also known as the first harmonic. It is the frequency at which the string vibrates in its simplest and most stable pattern.

How do you find the lowest resonant frequency on a string?

To find the lowest resonant frequency on a string, you can use the equation f = v/2L, where f is the frequency, v is the speed of the wave, and L is the length of the string. You can also use a string resonator or a tuning fork to determine the frequency by adjusting the length of the string until it produces a clear and loud sound.

What factors affect the lowest resonant frequency of a string?

The lowest resonant frequency of a string is affected by several factors, including the tension, length, and mass of the string. The type and thickness of the string material and the environment in which the string is vibrating can also impact the frequency.

Why is it important to find the lowest resonant frequency on a string?

Finding the lowest resonant frequency on a string is important because it helps determine the natural frequency of the string, which affects its sound and pitch. This information is crucial for musicians and scientists studying sound and vibrations.

Can the lowest resonant frequency on a string be changed?

Yes, the lowest resonant frequency on a string can be changed by altering the tension, length, or mass of the string. Changing the material of the string or the environment in which it is vibrating can also impact the frequency. Additionally, plucking or striking the string with different amounts of force can alter the lowest resonant frequency.

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