2. A plot of ln v vs. ln T should be a straight line. Why?

What is the slope of the straight line?

What is the intercept of the straight line?

1. A string is under a tension of 750.0 N. A 1.7 m length of the string has a mass of 4.8 grams. What is the speed of a transverse wave of wavelength 0.50 m in this string? What is the frequency of the wave?

2. A plot of ln v vs. ln T should be a straight line. Why?

What is the slope of the straight line?

What is the intercept of the straight line?

2. A plot of ln v vs. ln T should be a straight line. Why?

What is the slope of the straight line?

What is the intercept of the straight line?

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Dimitri S. | LightSpeed Ivy League TutorLightSpeed Ivy League Tutor

The speed of a Transverse wave under tension, fixed at both points is simply:

V= ((Tension (N) / (mass per unit length))^^{(1/2)}

Giving, Sqrt (750 / (.0048 kg / 1.7M))=

v (m/s) = SQRT (750/ 2.823x10^{-3) = 515 m/s}

Now, you guys out there, the frequency or the wavelength has NOTHING to do with the Physical set up above. That is an external excitation that CREATES the Frequency and the resulting wavelength.

f*Lambda = v=> f= v/lamba = 515 m/s / .5 = 103 Hz

The Slope is + 1/2 and the intercept is -1/2* ln (m/l)

Take the natural log of both sides of the V= SQRT (T/m/l) equation.

You will get a linear function ln(v) = 1/2 ln(T) - 1/2* ln (m/l)

Which in in the form of a linear equation:

y = mx + B

ln(v) = 1/2 ln (T) - ln (m/l)

LightSpeed Dimitri, PE

Meg,

If we let T be the tension (750 N) and μ be the linear density (1.7m/4.8*10^{-3}kg) then the wave speed is

v=(√T/μ)

The frequency is f=v/λ where λ is the wave length (.5 m)

the ln v= 1/2(ln T - ln μ) this is the equation of a straight line with slope = 1/2 and intercept==(1/2)*ln μ

You can do the arithmetic ok?

Regards

Jim

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