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A cord of mass 0.65 kg is stretched between two supports 8.0 m apart. If the tension in the cord is 120 N, how long will it take a pulse to travel from one support to the other?

Short Answer

Expert verified

The pulse takes a time of \(0.21\;{\rm{s}}\) to travel from one support to another.

Step by step solution

01

Understanding the speed of transverse wave

Whenever a wire string of specific length fix between two supports and a small disturbance create, then there would be propagation of a wave. The velocity of the wave can obtain with the help of tension of string, mass, and length of the string.

02

Identification of given data  

The given data can be listed below as,

  • The mass of a cord is,\(m = 0.65\;{\rm{kg}}\).
  • The distance between two supports is,\(l = 8.0\;{\rm{m}}\).
  • The tension in the cord is,\(T = 120\;{\rm{N}}\).
03

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, \(\mu \) is the mass per unit length and \(l\) is the length of the string.

04

Determining the required time taken by the pulse

The expression of the time taken by the pulse to travel from one support to other is given by,

\(\begin{aligned}{c}t &= \frac{l}{v}\\ &= \frac{l}{{\sqrt {\frac{{Tl}}{m}} }}\end{aligned}\)

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.

\(\begin{aligned}{c}v &= \frac{{\left( {8.0\;{\rm{m}}} \right)}}{{\sqrt {\frac{{\left( {120\;{\rm{N}}} \right)\left( {8.0\;{\rm{m}}} \right)}}{{0.65\;{\rm{kg}}}}} }}\\ &\approx 0.21\;{\rm{s}}\end{aligned}\)

Therefore, the time take by the pulse to move from one support to other is \(0.21\;{\rm{s}}\).

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Most popular questions from this chapter

Consider a wave traveling down a cord and the transverse motion of a small piece of the cord. Which of the following is true?

(a) The speed of the wave must be the same as the speed of a small piece of the cord.

(b) The frequency of the wave must be the same as the frequency of a small piece of the cord.

(c) The amplitude of the wave must be the same as the amplitude of a small piece of the cord.

(d) All of the above are true.

(e) Both (b) and (c) are true.

(III) A 1.60-kg object oscillates at the end of a vertically hanging light spring once every 0.45 s. (a) Write down the equation giving its position y (+upward) as a function of time t. Assume the object started by being compressed 16 cm from the equilibrium position (where y = 0), and released. (b) How long will it take to get to the equilibrium position for the first time? (c) What will be its maximum speed? (d) What will be the object鈥檚 maximum acceleration, and where will it first be attained?

P and S waves from an earthquake travel at different speeds, and this difference helps locate the earthquake 鈥渆picenter鈥 (where the disturbance took place). (a) Assuming typical speeds of and for P and S waves, respectively, how far away did an earthquake occur if a particular seismic station detects the arrival of these two types of waves 1.5 min apart? (b) Is one seismic station sufficient to determine the position of the epicenter? Explain.

Agent Arlene devised the following method of measuring the muzzle velocity of a rifle (Fig. 11鈥52). She fires a bullet into a 4.148-kg wooden block resting on a smooth surface, and attached to a spring of spring constant k= 162.7 N/m. The bullet, whose mass is 7.870 g, remains embedded in the wooden block. She measures the maximum distance that the block compresses the spring to be 9.460 cm. What is the speed v of the bullet?

A sailor strikes the side of his ship just below the surface of the sea. He hears the echo of the wave reflected from the ocean floor directly below 2.4 s later. How deep is the ocean at this point?

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