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The speed of a wave pulse on a string depends on the tension, F, in the string and the

mass per unit length, µ, of the string. Tension has SI units of kg • m • s-2 and the mass

per unit length has SI units of kg • m-1. What combination of F and µ must the speed of

the wave be proportional to?

Verified Answer :

Speed of a wave is in m * s-1, so we can figure this out through process of elimination to make the units of Tension and mass/length equal our goal units of m/s.

First, we can see there are no kg in our result, so no matter what we have to have F/μ or μ/F so we get our kg to cancel out.
Now we are left with either: 

\(\frac{F}{u} = \frac{m}{s^{2} } * \frac{1}{\frac{1}{m}} = \frac{m}{s^{2} } * \frac{m}{1} = \frac{m^{2}}{s^{2}}\)/F

or

 \(\frac{u}{F} = \frac{1}{m } * \frac{1}{\frac{m}{s^{2} }} = \frac{1}{m} * \frac{s^{2}}{m } = \frac{s^{2}}{m^{2}}\)

Now if our goal is m/s, the former of F/μ is closer, but off by a square. If we correct that by squaring F/μ we get

\(\sqrt{ \frac{F}{u} } = \sqrt{ \frac{ \frac{kg*m}{ s^{2} } }{ \frac{kg}{m} } } = \sqrt{ \frac{ m^{2} }{ s^{2} } } = \frac{m}{s} = v\)

Tension is the force conducted along the string and is measured in newton's, N.

The speed of wave is proportional to  \(\sqrt{\frac{F}{u} }\)

the ratio mass/length is read mass per unit length and represents the linear mass density of the string. This quantity is measured in kilograms/meter.

Tension is the force conducted along the string and is measured in newton's, N. The maximum tension that a string can withstand is called its tensile strength.

Since, The speed of a wave pulse on a string depends on the tension, F, in the string and the  mass per unit length, µ, of the string.

Speed of wave = \(\sqrt{\frac{Tension}{mass/length} }\)

Speed of wave=\(\sqrt{\frac{F}{u} }\)

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