Question

What will be the speed of these waves (in terms of V) if we increase M by a factor of 18.0, which stretches the wire to double its original length

Answer 1

The speed of the chord wave allows finding a new speed when the mass is increased and the chord length is:

• The velocity is: v = $\frac{v_o}{3}$  

A wave is a periodic movement of the particles that carries energy, but not matter, the speed of the waves is related to the properties of the medium by the relationship.

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

Where v is the speed of the wave, T the force and μ is the linear density.

Indicates that the applied mass increases by a factor of 18.0 and the length of the head is increased to twice the original.

The linear density of the cable is

          $\mu = \frac{m}{l}$

Where m is the mass of the cable and l is the length.

Let's use the subscript "o" for the initial conditions.

          M = 18.0 m₀

          l = 2 l₀

We look for the density.

         $\mu = \frac{18.0 m_o}{2.0 l_o}$mu = 18.0 mo / 2 lo

         $\mu = 9.0 \mu_o$  

We substitute in the expression for the velocity assuming that the tension is kept constant.

        $v= \sqrt{\frac{T}{9.0 \mu} }$  

        $v= \frac{v_o}{3}$  

In conclusion, using the speed of the chord wave we can find a new speed when the mass is increased and the chord length is:

• The velocity is: v = $\frac{v_o}{3}$

Learn more about wave speed here:  brainly.com/question/16824509