Question

A wave pulse travels down a slinky. The mass of the slinky is m = 0.87 kg and is initially stretched to a length L = 6.8 m. The wave pulse has an amplitude of A = 0.23 m and takes t = 0.478 s to travel down the stretched length of the slinky. The frequency of the wave pulse is f = 0.48 Hz.
1) What is the speed of the wave pulse? m/s2) What is the tension in the slinky?N3) What is the wavelength of the wave pulse?mPLEASE, CLEARLY WORK IT OUT CORRECTLY..THNKS

Answer 1

Answer:

1. $v=14.2259\ m.s^{-1}$

2. $F_T=25.8924\ N$

3. $\lambda=29.6373\ m$

Explanation:

Given:

• mass of slinky, $m=0.87\ kg$

• length of slinky, $L=6.8\ m$

• amplitude of wave pulse, $A=0.23\ m$

• time taken by the wave pulse to travel down the length, $t=0.478\ s$

• frequency of wave pulse, $f=0.48\ Hz=0.48\ s^{-1}$

1.

$\rm Speed\ of\ wave\ pulse=Length\ of\ slinky\div time\ taken\ by\ the\ wave\ to\ travel$

$v=\frac{6.8}{0.478}$

$v=14.2259\ m.s^{-1}$

2.

Now, we find the linear mass density of the slinky.

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

$\mu=\frac{0.87}{6.8}\ kg.m^{-1}$

We have the relation involving the tension force as:

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

$14.2259=\sqrt{\frac{F_T}{\frac{0.87}{6.8}} }$

$202.3774=F_T\times \frac{6.8}{0.87}$

$F_T=25.8924\ N$

3.

We have the relation for wavelength as:

$\lambda=\frac{v}{f}$

$\lambda=\frac{14.2259}{0.48}$

$\lambda=29.6373\ m$