A simple harmonic oscillator of angular frequ | vedclass

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Question

(C) The equation of motion for a forced harmonic oscillator is given by $m \frac{d^2x}{dt^2} + m \omega^2 x = F(t)$.Given $\omega = 2 \, rad \, s^{-1}

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$\sin(t) - \frac{1}{2} \sin(2t)$

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(C) The equation of motion for a forced harmonic oscillator is given by $m \frac{d^2x}{dt^2} + m \omega^2 x = F(t)$.Given $\omega = 2 \, rad \, s^{-1}$ and $F(t) = \sin(t)$,we have $\frac{d^2x}{dt^2} + 4x = \frac{1}{m} \sin(t)$.Assuming $m = 1$ for proportionality,the particular solution is of the form $x_p = A \sin(t)$.Substituting into the differential equation: $-A \sin(t) + 4A \sin(t) = \sin(t) \Rightarrow 3A = 1 \Rightarrow A = 1/3$.However,considering the initial conditions $x(0) = 0$ and $v(0) = 0$,the general solution is $x(t) = c_1 \cos(2t) + c_2 \sin(2t) + \frac{1}{3} \sin(t)$.Applying $x(0) = 0 \Rightarrow c_1 = 0$.Applying $v(0) = 0 \Rightarrow v(t) = 2c_2 \cos(2t) + \frac{1}{3} \cos(t) \Rightarrow 2c_2 + 1/3 = 0 \Rightarrow c_2 = -1/6$.Thus,$x(t) = \frac{1}{3} \sin(t) - \frac{1}{6} \sin(2t) = \frac{1}{3} (\sin(t) - \frac{1}{2} \sin(2t))$.Therefore,the position is proportional to $\sin(t) - \frac{1}{2} \sin(2t)$.

$A$ simple harmonic oscillator of angular frequency $\omega = 2 \, rad \, s^{-1}$ is acted upon by an external force $F = \sin(t) \, N$. If the oscillator is at rest in its equilibrium position at $t = 0$,its position at later times is proportional to

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