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Wize University Physics Textbook (Master) > Waves: Mechanical

Speed of Waves in Strings

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Speed of Waves in Strings


The most intuitive way of understanding a wave is a mechanical disturbance in a string. Think of a violin or guitar string, and how it vibrates when it is struck or plucked!



The physical properties of the string will affect the waves that travel through it:
  • The linear mass density (the mass per length).
μ=mL\boxed{\mu=\dfrac{m}{L}}
  • The tension in the string FTF_T.

Wave speed through a string or a rope depends on the linear mass density of the string (μ\mu) and the tension in the string (FT) as:


v =Fτμ\boxed{v\ =\sqrt{\dfrac{F_{\tau}}{\mu}}}


Wize Tip
The tighter the string, the more a point on the string is pulled by the next point, making it respond quicker. So the speed will increase with tension.

The heavier the string is, the more difficult it is to move a point on the string. So speed will decrease with increasing linear density.

Watch Out!
As you can see, speed of a wave only depends on the physical properties of the medium at which it is propagating. Hence, wave speed remains constant as long as we have the same string with the same tension no matter what is the frequency or wavelength of the wave.


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Example: Adjusting Tension in a String


A force is applied to a 10.0 m long rope with mass of 1.20 kg. The applied tension in the rope is 52.0 N

a) What tension is required to make the wave speed 3 times greater?
b) How long does it take for the wave to travel from one end of the rope to the other with the original tension on the rope?


a) Tension

The speed is given by: v=FTμv=\sqrt{\dfrac{F_T}{\mu}}

The new speed is 3 times greater:

v′=3vv'=3v

FT ′μ =3 FTμ \sqrt{\dfrac{F_T\ '}{\mu}}\ =3 \ \sqrt{\dfrac{F_T}{\mu\ }}

Square both sides to get:

FT ′=9⋅FTF_T\ '=9\cdot F_T
=9⋅52=9\cdot52
=468=468 (N)

b) Time of travel

First, find the wave speed:

v=FTμ=FTmL =FTLm v=\sqrt{\dfrac{F_T}{\mu}}=\sqrt{\dfrac{F_T}{\frac{m}{L}}}\ =\sqrt{\dfrac{F_TL}{m}\ }
=52(10)1.2 =\sqrt{\dfrac{52\left(10\right)}{1.2}\ }
=20.8=20.8 (m/s)
Now we can find the time:

t=xv=1020.8=0.480t=\dfrac{x}{v}=\dfrac{10}{20.8}=0.480 (s)

Practice: String Mass Density


A horizontal string has its tension measured to be 1 N1\ \mathrm{N}. A sinusoidal wave is sent down the string, its equation is found to be
y(x,t)=(15 cm)sin⁡(kx−ωt+ϕ)y(x,t) = (15\ \mathrm{cm}) \sin(kx-\omega t+\phi)
with k=0.2 rad/cmk = 0.2\ \mathrm{rad}/\mathrm{cm}, and is measured to have a period of oscillation of T=0.5 sT = 0.5\ \mathrm{s}.

What is the linear mass density of the string?