Question

Two children are sending signals along a cord of total mass 0.50 kg tied between tin cans with a tension of 35 N. It takes the vibrations in the string 0.55 s to go from one child to the other. How far apart are the children?

Short answer

The distance between the two children is\(21.12\;{\rm{m}}\).

Step-by-step solution

Understanding about the wave velocity

Whenever a wave moves in a specific length of wire, the value of the speed of the wave depends on the various parameters like wire’s tension, length, mass, etc.The wave speed in the wire has a relation to the square root of the wire’s tension.

Identification of given data

The given data can be listed below as,

• The total mass of cord is\(m = 0.5\;{\rm{kg}}\).
• The value of tension in the cord is\(T = 35\;{\rm{N}}\).
• The time taken for vibrations is\(t = 0.55\;{\rm{s}}\).

Defining the expression of wave speed

The expression for the speed of the transverse waves in the cord is,

\(\begin{aligned}{c}v &= \sqrt {\frac{T}{\mu }} \\ &= \sqrt {\frac{T}{{\left( {\frac{m}{L}} \right)}}} \\ &= \sqrt {\frac{{TL}}{m}} \end{aligned}\)

Here, \(v\) is the wave speed, \(\mu \) is the mass per unit length and \(L\) is the distance between the two children.

Calculate the distance between the children

From the above expression, the equation of the distance between the two children is given by,

\(\begin{aligned}{c}v &= \sqrt {\frac{{TL}}{m}} \\\left( {\frac{L}{t}} \right) &= \sqrt {\frac{{TL}}{m}} \\\left( {\frac{L}{{{t^2}}}} \right) &= \left( {\frac{T}{m}} \right)\\L &= \left( {\frac{{T{t^2}}}{m}} \right)\end{aligned}\)

Substituteall the known values in the above equation.

\(\begin{aligned}{c}L &= \frac{{\left( {35\;{\rm{N}}} \right){{\left( {0.55\;{\rm{s}}} \right)}^2}}}{{\left( {0.5\;{\rm{kg}}} \right)}}\\ &\approx 21.12\;{\rm{m}}\end{aligned}\)

Therefore, the distance between the two children is\(21.12\;{\rm{m}}\).