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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

01

Write the given information 

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

02

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
λn=2Ln

Here n=3,
λ3=2L3

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=1λTµ

Substitute the values
f=32Lk(0.08)µ......(1)

Now, to get two antinodes, n=2, wavelength λ2is
λ2=2L2=L

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

Now the frequency of the wave in this case is
f=1Lk∆x’µ......(2)

Compare both the equations
32Lk(0.08)µ=1Lk(∆x’)µ320.08=(∆x’)9(0.08)=4(∆x’)∆x’=18cm
∆x’= 18 cm

Thus the stretch in the wire is 18cm

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