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Classical Mechanics Level 5

A thin conducting wire of length L L , resistance R R made of a non-magnetic material of uniform linear mass density μ \mu is fitted between two metal walls which are connected to each other via a thick metal strip. A tension T T is maintained in the wire.

The wire is made to vibrate in its fundamental mode. At an instant when the wire is at one of the extreme positions, a uniform magnetic field of magnitude B 0 B_0 perpendicular to the plane of vibration is turned on in the area between the walls. This magnetic field damps the vibration of the wire.

Find the time taken(in seconds) for the amplitude of vibration to become half its initial value.

Details and Assumptions:

  1. L = 1 m L=1 m , μ = 0.25 K g m − 1 \mu=0.25Kgm^{-1} , T = 100 N T=100N , R = 10 Ω R=10\Omega , B 0 = 0.1 π T B_0=0.1\pi T

  2. Assume tension to be uniform throughout the string

  3. Assume that the wave oscillates in fundamental mode throughout the experiment.

You may use any computer resource to solve the problem. If you genuinely feel that this is incorrect, feel free to report.


The answer is 43.323.

Raghav Vaidyanathan by Raghav Vaidyanathan , 23, USA

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2 solutions

Jatin Yadav
Apr 27, 2015

First, we find the total energy of the wire.

When it is in mean position, potential energy =0.

At a distance x,

∣ v ∣ = 2 A ω sin ⁡ ( ω t ) sin ⁡ ( k x ) |v|= 2A\omega\sin(\omega t) \sin(kx)

Clearly, asis it at mean position, cos ⁡ ( ω t ) = 0 \cos(\omega t) = 0

Now, kinetic energy = total energy = 1 2 ∫ 0 π / k μ ( 2 A ω sin ⁡ ( k x ) ) 2 d x \dfrac{1}{2} \displaystyle \int_{0}^{\pi/k} \mu (2A \omega \sin(kx))^2 {\mathrm dx}

We get total energy as E = μ A 2 ω 2 L E = \mu A^2 \omega^2 L

Now, induced voltage,

V = ∫ 0 L 2 A ω sin ⁡ ω t sin ⁡ ( π L x ) B d x V = \displaystyle \int_{0}^{L} 2 A\omega \sin \omega t \sin \bigg(\dfrac{\pi}{L} x\bigg) B dx

= 4 A B ω L π sin ⁡ ω t \dfrac{4AB \omega L}{\pi} \sin \omega t

Hence, energy dissipated per second = V 2 R \dfrac{V^2}{R}

Also, this energy comes from loss of energy from wire, which equals − 2 μ ω 2 L A d A d t -2 \mu \omega^2 L A \dfrac{dA}{dt}

Hence, − 2 μ ω 2 L d A d t = ( 4 A B ω L π ) 2 sin ⁡ 2 ω t R -2 \mu \omega^2 L \dfrac{dA}{dt} = \Bigg(\dfrac{4 A B \omega L}{\pi}\Bigg)^2 \dfrac{\sin^2 \omega t}{R}

Hence, d A A = − 8 B 2 L π 2 μ R sin ⁡ 2 ω t d t \dfrac{dA}{A} = -\dfrac{8 B^2 L}{\pi^2 \mu R} \sin^2 \omega t dt

Hence,we get t 2 − sin ⁡ 2 ω t 4 ω = π 2 μ R ln ⁡ 2 8 B 2 L \dfrac{t}{2} - \dfrac{\sin 2 \omega t}{4 \omega} = \dfrac{\pi^2 \mu R \ln 2}{8 B^2 L}

ω = 2 π f = 2 π v λ = π L T μ = 20 π \omega = 2 \pi f = 2 \pi \dfrac{v}{\lambda} = \dfrac{\pi}{L} \sqrt{\dfrac{T}{\mu}} = 20 \pi

Using this value, we get t = 43.323 t = 43.323

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This is how I did it too. +1

Raghav Vaidyanathan - 6 years, 1 month ago

The amplitude of the wave as a function of time is given by:

a ( t ) = a 0 e − K ( t ) a(t)={ a }_{ 0 }{ e }^{ -K(t) }

K ( t ) = B 0 2 λ π 2 R μ ω [ 2 ω t − sin ⁡ ( 2 ω t ) ] K(t)=\frac { { { B }_{ 0 } }^{ 2 }\lambda }{ \pi^2 R\mu \omega } \left[ 2\omega t-\sin { (2\omega t) } \right]

λ = 2 L , ω = π L T μ \lambda =2L,\quad \omega =\frac { \pi }{ L } \sqrt { \frac { T }{ \mu } }

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