Question

A steel wire is used to stretch the spring of FIGURE P17.42. An oscillating magnetic field drives the steel wire back and forth. A standing wave with three antinodes is created when the spring is stretched 8.0 cm. What stretch of the spring produces a standing wave with two antinodes?

Short answer

The stretch in the wire is 18cm

Step-by-step solution

Write the given information 

The stretch in wire, ∆x= 8 cm =0.08 m
The three antinodes are produced, n=3

To determine the stretch in the wire when n=2

Let the wavelength is denoted by λ
The wavelength of the nth harmonic frequency is given by
${\lambda }_n=\frac{2L}{n}$

Here n=3,
${\lambda }_3=\frac{2L}{3}$

Let the tension in the string is denoted by T
The expression of tension is given by T= k∆x
Here, k is the spring constant for wire.

Now, the fundamental frequency of the wave is given by

role="math" localid="1650110684089" $f=\left(\frac{1}{\lambda }\right)\sqrt{\frac{T}{\mu }}$

Substitute the values
$f=\left(\frac{3}{2L}\right)\sqrt{\frac{k(0.08)}{\mu }}......\left(1\right)$

Now, to get two antinodes, n=2, wavelength λ₂is
${\lambda }_2=\frac{2L}{2}=L$

The stretch in the wire for this case is ∆x’

Now the frequency of the wave in this case is
$f=\frac{1}{L}\sqrt{\frac{k\left(∆x’\right)}{\mu }}......\left(2\right)$

Compare both the equations
$$\begin{gathered} \left(\frac{3}{2L}\right)\sqrt{\frac{k(0.08)}{\mu }}=\frac{1}{L}\sqrt{\frac{k(∆x’)}{\mu }} \\ \frac{3}{2}\sqrt{0.08}=\sqrt{(∆x’)} \\ 9(0.08)=4(∆x’) \\ ∆x’=18cm \end{gathered}$$
∆x’= 18 cm

Thus the stretch in the wire is 18cm