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11-41P

Expert-verifiedFound in: Page 292

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\(4.82\;{\rm{N}}\).

**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.

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, \(t = 0.85\;{\rm{s}}\).

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.

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\(4.82\;{\rm{N}}\).

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