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Found in: Page 292

### Physics Principles with Applications

Book edition 7th
Author(s) Douglas C. Giancoli
Pages 978 pages
ISBN 978-0321625922

# A 0.40-kg cord is stretched between two supports, 8.7 m apart. When one support is struck by a hammer, a transverse wave travels down the cord and reaches the other support in 0.85 s. What is the tension in the cord?

The tension in the cord is$$4.82\;{\rm{N}}$$.

See the step by step solution

### Step by Step Solution

Whenever a specific length of wire/rod stretches with the help of an external force, then there would be a tension generated in the wire/rod. The value of tension force would be the same at each section of the wire/rod.

## Identification of given data

The given data can be listed below as,

• The mass of a cord is, $$m = 0.40\;{\rm{kg}}$$.
• The distance between two supports is, $$l = 8.7\;{\rm{m}}$$.
• The time taken by the transverse wave is, $$t = 0.85\;{\rm{s}}$$.

## Defining the expression of the speed of transverse wave

The speed of the transverse wave in the string can be expressed as,

\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 speed of the transverse wave in the string, $$T$$ is the tension in the cord, $$\mu$$ is the mass per unit length and $$l$$ is the length of the string.

## Determining the tension in the cord

The expression of the velocity of transverse wave is given by,

\begin{aligned}{c}v &= \frac{l}{t}\\\sqrt {\frac{{Tl}}{m}} &= \frac{l}{t}\\T &= \frac{{ml}}{{{t^2}}}\end{aligned}

Here, $$t$$ is the time taken by the pulse to travel from one support to other.

Substitute all the known values in the above equation.

\begin{aligned}{c}T &= \frac{{\left( {0.40\;{\rm{kg}}} \right)\left( {8.7\;{\rm{m}}} \right)}}{{{{\left( {0.85\;{\rm{s}}} \right)}^2}}}\\ &\approx 4.82\;{\rm{N}}\end{aligned}

Therefore, the tension in the cord is$$4.82\;{\rm{N}}$$.