The E string of a violin has a linear density of 0.5 g / m and is subjected to a

80N of tension, tuned for a frequency = 660 Hz
a) What is the length of the rope?

Oscillations? wavelength=speedsound/freeq

so what is the speed of sound on the string? The speed of propagation of a wave in a string (v) is proportional to the square root of the tension of the string (F) (discovered by Vincenzo Galilei in the late 1500s) and inversely proportional to the square root of the linear density (\mu) of the string:

speed= sqrt(tension/mu) remember mu needs to be in kg/meter

so now you have speed, calculate wavelength, then the length of the rope is one half wavelength...

To find the length of the E string, we need to use the formula for the speed of a wave on a string:

v = √(T/μ)

Where:
v = speed of the wave
T = tension in the string
μ = linear mass density of the string

In this case, we are given the tension (T = 80N) and the linear mass density (μ = 0.5 g/m). However, before we proceed, let's convert the linear mass density to kg/m to maintain consistent units:

μ = 0.5 g/m = 0.5 × 10^(-3) kg/m

Now, let's substitute these values into the formula and solve for the speed of the wave:

v = √(80N / 0.5 × 10^(-3) kg/m)
v = √(80 / 0.5 × 10^(3))
v ≈ 28.28 m/s

The speed of the wave is approximately 28.28 m/s.

Next, to find the length of the rope (string), we can use the formula for the frequency of a wave on a string:

f = (v / λ)

Where:
f = frequency of the wave
v = speed of the wave
λ = wavelength of the wave

In this case, we know the frequency (f = 660 Hz), and we need to find the wavelength (λ) to get the length of the string. Rearranging the formula, we have:

λ = v / f
λ = 28.28 m/s / 660 Hz
λ ≈ 0.043 m

The wavelength is approximately 0.043 m.

Finally, since the length of the string is equal to half of the wavelength for a standing wave, we can calculate the length of the E string:

Length of string = λ / 2
Length of string = 0.043 m / 2
Length of string ≈ 0.0215 m

The length of the E string is approximately 0.0215 m.