What Is the Maximum Velocity of a Particle on a Tensioned String?

[oldunion]

A string that is under 46.0 N of tension has linear density 5.00 g/m. A sinusoidal wave with amplitude 2.80 cm and wavelength 1.80 m travels along the string.
what is the maximum velocity of a particle on the string.

i don't know how to solve this without a frequency or length of string.

 

[OlderDan]

I don't recall the exact relationship, but from density and tension you can find the velocity. Since you have wavelength, you can get the frequency. I'm sure you have the relationship in your text or reference material.

From googling

c = sqrt(T/density per length)

 

[oldunion]

wavelength= velocty/frequency.
and i don't understand your reference for density and tension

 

[jtbell]

i don't understand your reference for density and tension

The speed of a wave in a stretched string depends on the tension in the string, and on the mass of the string per unit length (kg/m) which is usually called the linear mass density. Rewriting OlderDan's formula a bit (because I don't like to use c for any speed other than the speed of light, and the linear mass density is usually called \mu):

v = \sqrt {\frac {T}{\mu}}

 

[ZapperZ]

Wait.. isn't this "v" the velocity of propagation? The original question asked for

what is the maximum velocity of a particle on the string.

This is the first derivative of the wave function at zero displacement, no?

Zz.

 

[HallsofIvy]

Do you understand that a "particle on the string" is moving up and down, not moving along the string with the wave?

 

[clive]

v=\sqrt{\frac{T}{\mu}}
\omega=2\pi \frac{v}{\lambda}
v_{max}=\omega A

 

[mickles]

v=\sqrt{\frac{T}{\mu}}
\omega=2\pi \frac{v}{\lambda}
v_{max}=\omega A

I understand the equations but why do all of them fit, why not just use the last Vmax to get the answer instead of subbing and letting the end equation be

I don't understand how getting this was the answer

Vmax = (2pi*A)/wavelength * sqrt(Tension/mu)