Question

A wave pulse travels along a string at a speed of 200 cm/s. What will be the speed if:

Answer 1

Answer:

a) $v_f = \sqrt{\frac{2TL}{m}} = \sqrt{2} v = \sqrt{2}2m/s =2.83 m/s$

The velocity increase by a factor of $\sqrt{2}$

b) $v_f = \sqrt{\frac{TL}{4m}} = \frac{1}{2} v = \frac{1}{2} *2m/s =1 m/s$

The velocity decrease by a factor of 2.

c) $v_f = \sqrt{\frac{4TL}{m}} = 2} v = 2 *2m/s =4m/s$

The velocity increase by a factor of 2

d) $v_f = \sqrt{\frac{4TL}{4m}} = v = 2m/s$

The velocity not changes.

Explanation:

For this case we know that the velocity is $v = 200 cm/s = 2m/s$

$v_f$ represent the final velocity after the changes specified,

Part a

The formula for the speed of a wave in a string is given by:

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

And the linear density is defined as:

$\rho = \frac{m}{L}$

And if we replace this we got:

$v = \sqrt{\frac{TL}{m}}$

If the tension mass is doubled we have this:

$v_f = \sqrt{\frac{2TL}{m}} = \sqrt{2} v = \sqrt{2}2m/s =2.83 m/s$

The velocity increase by a factor of $\sqrt{2}$

Part b

If we mass is quadrupled we have this:

$v_f = \sqrt{\frac{TL}{4m}} = \frac{1}{2} v = \frac{1}{2} *2m/s =1 m/s$

The velocity decrease by a factor of 2.

Part c

If the length is quadrupled we have this:

$v_f = \sqrt{\frac{4TL}{m}} = 2} v = 2 *2m/s =4m/s$

The velocity increase by a factor of 2

Part d

For this case we know that the mass and the length are both quadrupled and we got:

$v_f = \sqrt{\frac{4TL}{4m}} = v = 2m/s$

The velocity not changes.