Question

Two children are sending signals along a cord of total mass 0.45 kg tied between tin cans with a tension of 33 N . It takes the vibrations in the string 0.51 s to go from one child to the other.a. How far apart are the children?b. Express your answer using two significant figures.

Answer 1

Answer:

$l\approx 19m$

Explanation:

Given:

Tension in the string, $F_T=33N$

mass of the string, $m=0.45kg$

time taken to transmit the sound, $t=0.51 s$

We have the relation of the given with velocity 'v' as:

$v^2=\frac{F_T}{\mu}$ ................................(1)

where:

${\mu}$= linear mass density of the string=$\frac{m}{l}$

we know:

$v=\frac{l}{t}$

where:

$l=$length of the string

Putting the corresponding values in the eq. (1)

$(\frac{l}{t})^2=\frac{F_T}{(\frac{m}{l}) }$

$l=\frac{F_T\times t^2}{m}$

$\Rightarrow l=\frac{33\times 0.51^2}{0.45}$

$l=19.074\,m$

$l\approx 19m$