Chapter 17: Q. 42 (page 485)

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

Expert verified

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
$λ_{n} = \frac{2 L}{n}$

Here n=3,
$λ_{3} = \frac{2 L}{3}$

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" $f = (\frac{1}{λ}) \sqrt{\frac{T}{µ}}$

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

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

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

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

Compare both the equations
$(\frac{3}{2 L}) \sqrt{\frac{k (0.08)}{µ}} = \frac{1}{L} \sqrt{\frac{k (∆ x ’)}{µ}} \frac{3}{2} \sqrt{0.08} = \sqrt{(∆ x ’)} 9 (0.08) = 4 (∆ x ’) ∆ x ’ = 18 c m$
∆x’= 18 cm