Derivation of the wave equation F=qE

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<ul><li><p>8/11/2019 Derivation of the wave equation F=qE</p><p> 1/3</p><p>Derivation of the wave equation.</p><p>The wave equation is derived for a very thin uniform string. the string has mass density and is in</p><p>equilibrium under tension T.</p><p>Now we displace the string slightly from its equilibrium position. And exaggerated view of a small</p><p>(infinitesimal) segment is shown below:</p><p>The length of the segment is x, hence its mass is:</p><p>m x (0.1)</p><p>The motion of the string is in the y direction. This means that the horizontal components of the</p><p>tensionR</p><p>T andL</p><p>T must be equal (and in opposite directions) to some constant :</p><p>cos cosL L R R</p><p>T T (0.2)</p><p>In the vertical direction we can write Newton's second law equation, namelyF manoting that</p><p>the acceleration is the second derivative of ywith respect to time:</p><p>2</p><p>2sin sin</p><p>y R R L L</p><p>yF T T m </p><p>t (0.3)</p><p>sin siny R R L L</p><p>F T T qE </p><p>The mass is given in equation (1.1) and we also divide equation (1.3) by :</p><p>2</p><p>2</p><p>sin sinR R L L</p><p>T T x y</p><p>t (0.4)</p><p>2</p><p>2</p><p>sin sinR R L L electron</p><p>T T V y</p><p>t</p><p>However, according to (1.2) we can write it as:</p></li><li><p>8/11/2019 Derivation of the wave equation F=qE</p><p> 2/3</p><p>2</p><p>2</p><p>2</p><p>2</p><p>2</p><p>2</p><p>sin sin</p><p>sin sin</p><p>sin sin</p><p>cos cos</p><p>R R L L</p><p>R R L L</p><p>R R L L</p><p>R R L L</p><p>T T x y</p><p>t</p><p>T T x y</p><p>t</p><p>T T x y</p><p>tT T</p><p>2</p><p>2tan tan</p><p>R L</p><p>x y</p><p>t (0.5)</p><p>2</p><p>2tan tan electron</p><p>R L</p><p>V y</p><p>t </p><p>But the tangents of the angles are, by definition, the derivatives of y with respect to x, hence:</p><p>2</p><p>2</p><p>x x x</p><p>y y x y </p><p>x x t (0.6)</p><p>Dividing both sides by x we get:</p><p>2</p><p>2</p><p>1</p><p>x x x</p><p>y y y</p><p>x x x t (0.7)</p><p>2</p><p>2</p><p>1</p><p>electron x x x</p><p>y y y</p><p>V x x t</p><p>If we take the limit of an infinitesimal segment, 0x , the left hand side becomes the second</p><p>partial derivative with respect to x(by definition).</p><p>Cant get double derivative here because ofelectron</p><p>V </p><p>Furthermore, when the segment is very short and the string is not far from equilibrium, the tension</p><p>is for all practical reasons equals the original tension T.</p><p>This turns equation (1.7) into the well known one dimensional wave equation:</p><p>2 2</p><p>2 2</p><p>y y</p><p>Tx t (0.8)</p><p>Or:</p></li><li><p>8/11/2019 Derivation of the wave equation F=qE</p><p> 3/3</p><p>2 2</p><p>2 2</p><p>T y y</p><p>Tx t (0.9)</p><p>The units of the constant coefficient T are:</p><p>-2 2</p><p>-1 2</p><p>kg m s m</p><p>kg m s</p><p>T (0.10)</p><p>This are units of the square of a speed.</p><p>Hence we can define the wave's propagation speed:</p><p>2 T</p><p>c (0.11)</p><p>And the one dimensional wave equation is:</p><p>2 22</p><p>2 2</p><p>y yc</p><p>x t (0.12)</p></li></ul>

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