/sci/ - Science & Math

>>11594460
Oops. Yeah, you're right about the magnitude of B at x and y, but consider than the integrals along the top and bottom of the rectangle are both not zero. In fact, they are both negative and have same magnitude. These two combine with the difference of the integrals along BC and DA to get zero.
>>>/wsr/809597
>>>/wsr/809620
>2
Draw a free body diagram and make sure to pick proper coordinates. Recall Hooke's law for springs. You should get the equation of motion for the mass: [math] F(t)-kx=m\ddot{x}\implies 4\ddot{x}+9x=\sin(3t/2) [/math]. Write down the characteristic equation and use the method of undetermined coefficients to get [eqn] x(t)=C_1\cos(3t/2)+C_2\sin(3t/2)-(1/12)t\cos(3t/2) [/eqn] Solve for the constants with initial conditions, C1=0, C2=1/18.
>3
It's just a separable ODE. Write down [eqn] \int2\text{ d}x=\int\frac{3\text{ d}y}{(1+y)(1-2y)}=\int\frac{1}{1+y}+\frac{2}{1-2y}\text{ d}y [/eqn] and compute.
>4
Part i is very easy. Part ii involves an integration factor. Recall that for [math] u'+f(x)u=g(x) [/math] you can let [math] \mu=\exp\int f(x)\text{ d}x [/math] then [math] u=\int\mu g(x)\text{ d}x/\mu [/math] and you finally end up with [math] u=2x+Cx^2 [/math]. You can easily get y now.

>>>/wsr/795140
[eqn] \mathbf{F}_{24}=k\frac{q_2q_4}{r_{24}^2}\hat{\mathbf{r}}_{14}=k\frac{q_2q_4}{(\sqrt{2}d)^2}\big[(\sqrt{2}/2)\hat{\mathbf{i}}-(\sqrt{2}/2)\hat{\mathbf{j}}\big]=\frac{k}{d^2}q_2q_4\big[(\sqrt{2}/4)\hat{\mathbf{i}}-(\sqrt{2}/4)\hat{\mathbf{j}}\big] \\
\mathbf{F}_{14}=...=\frac{k}{d^2}(q_1q_4/4)\hat{\mathbf{i}}\\ \mathbf{F}_{34}=...=\frac{k}{d^2}q_3q_4\hat{\mathbf{i}} [/eqn]
so
[eqn]\sum\mathbf{F}=\frac{k}{d^2}\Bigg[\bigg(\frac{\sqrt{2}q_2q_4}{4}+\frac{q_1q_4}{4}+q_3q_4\bigg)\hat{\mathbf{i}}-\bigg(\frac{\sqrt{2} q_2q_4}{4}\bigg)\hat{\mathbf{j}}\Bigg][/eqn]
you do the rest

>>/wsr/795140
[eqn] \mathbf{F}_{24}=k\frac{q_2q_4}{r_{24}^2}\hat{\mathbf{r}}_{14}=k\frac{q_2q_4}{(\sqrt{2}d)^2}\big[(\sqrt{2}/2)\hat{\mathbf{i}}-(\sqrt{2}/2)\hat{\mathbf{j}}\big]=\frac{k}{d^2}q_2q_4\big[(\sqrt{2}/4)\hat{\mathbf{i}}-(\sqrt{2}/4)\hat{\mathbf{j}}\big] \\
\mathbf{F}_{14}=...=\frac{k}{d^2}(q_1q_4/4)\hat{\mathbf{i}}\\ \mathbf{F}_{34}=...=\frac{k}{d^2}q_3q_4\hat{\mathbf{i}} [/eqn]
so
[eqn]\sum\mathbf{F}=\frac{k}{d^2}\Bigg[\bigg(\frac{\sqrt{2}q_2q_4}{4}+\frac{q_1q_4}{4}+q_3q_4\bigg)\hat{\mathbf{i}}-\bigg(\frac{\sqrt{2} q_2q_4}{4}\bigg)\hat{\mathbf{j}}\Bigg][/eqn]
you do the rest