Question

A wire with mass \(40.0\;{\rm{g}}\) is stretched so that its ends are tied down at points \(80.0\;{\rm{cm}}\) apart. The wire vibrates in its fundamental mode with frequency \(60.0\;{\rm{Hz}}\) and with an amplitude at the antinodes of \(0.300\;{\rm{cm}}\).

(a) What is the speed of propagation of transverse waves in the wire?

(b) Compute the tension in the wire.

(c) Find the maximum transverse velocity and acceleration of particles in the wire.

Short answer

(a) The speed of propagation of transverse waves in the wire is, \(96.0\;{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. \\} {\rm{s}}}\).

Step-by-step solution

Identification of the given data

The given data can be listed below as,

• The length of wire is, \(L = 80.0\;{\rm{cm}} = 0.80\;{\rm{m}}\).
• The frequency is, \(f = 60\;{\rm{Hz}}\).

Concept

To determine the velocity, first apply the equation for wave velocity and substitute fundamental wavelength in terms of wire length.

The relation between frequency, wavelength, and speed is expressed as,

\(f=\frac{v}{\lambda}\) …(2)

Here\(\lambda \)is the wavelength,\(f\)is the frequency, and\(v\)is the speed of wave.

Determination of the speed of propagation of transverse waves in the wire

The fundamental standing wave is illustrated in the below figure,

From the figure,

\(\begin{array}{c}\frac{\lambda }{2} = L\\\lambda = 2L\end{array}\)

Substitute \(0.80\;{\rm{m}}\) for \(L\) in the above equation.

\(\begin{array}{c}\lambda=2\times0.80\;{\rm{m}}\\\lambda=1.60\;{\rm{m}}\end{array}\)

The wavelength is, \(\lambda = 1.60\;{\rm{m}}\).

Substitute \(1.60\;{\rm{m}}\) for \(\lambda \) and \(60\;{\rm{Hz}}\) for \(f\) in the equation (1).

\(\begin{array}{c}60\;{\rm{Hz}} = \frac{v}{{1.60\;{\rm{m}}}}\\v = 60\;{\rm{Hz}} \times 1.60\;{\rm{m}}\\v = 96.0\;{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. \\} {\rm{s}}}\end{array}\)

Hence the speed of propagation of transverse waves in the wire is, \(96.0\;{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. \\} {\rm{s}}}\).