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klio [65]
3 years ago
9

A violin string is 45.0 cm long and has a mass of 0.242 g. When tightened on the neck of the violin, the distance between the pi

n and fret is 42.5 cm. If the string is tightened to play a 'Concert A' of 440. Hz as its fundamental/ first harmonic, what is the tension in the violin string
Physics
1 answer:
stiks02 [169]3 years ago
3 0

Answer:

The tension is 75.22 Newtons

Explanation:

The velocity of a wave on a rope is:

v=\sqrt{\frac{TL}{M}} (1)

With T the tension, L the length of the string and M its mass.

Another more general expression for the velocity of a wave is the product of the wavelength (λ) and the frequency (f) of the wave:

v= \lambda f (2)

We can equate expression (1) and (2):

\sqrt{\frac{TL}{M}}=\lambda f

Solving for T

T= \frac{M(\lambda f)^2}{L} (3)

For this expression we already know M, f, and L. And indirectly we already know λ too. On a string fixed at its extremes we have standing waves ant the equation of the wavelength in function the number of the harmonic N_{harmonic} is:

\lambda_{harmonic}=\frac{2l}{N_{harmonic}}

It's is important to note that in our case L the length of the string is different from l the distance between the pin and fret to produce a Concert A, so for the first harmonic:

\lambda_{1}=\frac{2(0.425m)}{1}=0.85 m

We can now find T on (3) using all the values we have:

T= \frac{2.42\times10^{-3}(0.85* 440)^2}{0.45}

T=75.22 N

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