Q76CP - University Physics with Modern Physics

Question

Longitudinal Waves on a Spring. A long spring such as a ${\mathbf{Slinky}}^{\mathbf{TM}}$ is often used to demonstrate longitudinal waves. (a) Show that if a spring that obeys Hooke’s law has mass m, length L, and force constant k′, the speed of longitudinal waves on the spring is  $\mathbf{v}\mathbf{=}\mathbf{L}\sqrt{\mathbf{k}\mathbf{\text{'}}}\mathbf{m}$

 (see Section 16.2). (b) Evaluate v for a spring with m = 0.250 kg, L = 2.00 m, and k′ = 1.50 N/m.

Step-by-Step Solution

Answer

a) We have made use of the principle that states that the impulse equals the change in the momentum to reach the result.  b) v =4.9m/s 

STEP 1 Concept of the change in the momentum

The change in the momentum is given as $\frac{\mathbf{m}}{\mathbf{L}\mathbf{ }}{\mathbf{v}}^{\mathbf{2}}\mathbf{t}$

STEP 2 Derive the change in the momentum

a) consider a longitudinal wave that travels through the slinky by comprising a part of the slinky with a length of (vt), where (v) is the speed of the pulse "the speed of the propagation of the wave". So, the mass of this part of the slinky is $\left(\mathrm{\mu vt}\right)$ , where (g) is the mass per unit length of the slinky. Hence, we can write the change in the longitudinal momentum as follows 

Change in momentum = (The mass of the compressed part) x The change in velocity

                                        = $\left(\mathrm{\mu vt}\right)\times \mathrm{V}$

Notice that, $\mathrm{\mu }=\frac{\mathrm{m}}{\mathrm{L}}$ where (m) is the mass of the slinky and (L) is the length of the slinky, and change in the momentum of the pulse between the first point of the slinky with zero velocity and the final point of the slinky with velocity (v). 

 Thus, Change in momentum = $\frac{\mathrm{m}}{\mathrm{L}}{\mathrm{v}}^2\mathrm{t}$

STEP 3 Concept of hooks law

In order to find the impulse for the wave pulse in the slinky we make use of the principle that states 

that the impulse equals the change in the momentum, so, we need first to find the force acting on the compressed part when it reaches the end of the slinky, that force can be determined using Hook's law 

In the case, k= k' and $∆\mathrm{X}$= L, so, the magnitude of F becomes 

    $\mathrm{F}=\mathrm{k}\text{'}\mathrm{L}$

$\mathrm{Impulse}=\mathrm{Ft}=\mathrm{k}\text{'}\mathrm{Lt}$

Equating the Impulse and the Change in momentum 

$\mathrm{k}\text{'}\mathrm{Lt}=\frac{\mathrm{m}}{\mathrm{L}}{\mathrm{v}}^2\mathrm{t}$

$$\begin{gathered} {\mathrm{v}}^2=\frac{\mathrm{k}\text{'}{\mathrm{L}}^2}{\mathrm{m}} \\ \mathrm{v}=\mathrm{L}\sqrt{\frac{\mathrm{k}}{\mathrm{m}}} \\ \mathrm{v}=\left(2\mathrm{m}\right)\sqrt{\frac{1.5\mathrm{N}/\mathrm{m}}{0.25\mathrm{kg}}} \\ \mathrm{v}=4.9\mathrm{m}/\mathrm{s} \end{gathered}$$

Therefore, the velocity is $4.9\mathrm{m}/\mathrm{s}$