Edits to “Statement”, “Solution”, “Answer”

Tete edited Aug 4, 2026, 3:27 AM

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revision #19820 parent #19819 ← older

en/3.5.36.md +8 −4

		@@ -1,13 +1,17 @@
			### Statement

		−	$3.5.36.$ [Insert the problem statement]
		+	$3.5.36.$ a. What is the necessary forcing force for an oscillator of mass $m$ with damping coefficient $\gamma$ to begin to perform harmonic oscillations with natural frequency $\omega_0$ according to the law $x=A\cos(\omega_0t-\phi)$?

		+	b. The amplitude of the forcing force is $F_0$, its frequency $\omega=\omega_0$. Determine the amplitude of the forced oscillations. How many times is it the deflection of the oscillator when a constant force $F_0$ is applied?
		+
			### Solution

		−	Substituting the particular (i.e. steady-state) solution $x=A\cos(\omega_0t-\phi)$ into the expression $\ddot x+2\gamma\dot x$+\omega_0^2x, we obtain $-2\gamma\omega A\sin(\omega_0t-\phi)$. (Note that the first and third terms cancel out, because the particular solution happens to be the solution to the homogeneous equation $\ddot x+\omega_0^2x=0$.) Thus, the required forcing is $F(t)=-2m\gamma\omega A\sin(\omega_0t-\phi)$.
		+	Substituting the particular (i.e. steady-state) solution $x=A\cos(\omega_0t-\phi)$ into the expression $\ddot x+2\gamma\dot x+\omega_0^2x$, we obtain $-2\gamma\omega_0A\sin(\omega_0t-\phi)$. (Note that the first and third terms cancel out, because the particular solution happens to be the solution to the homogeneous equation $\ddot x+\omega_0^2x=0$.) Thus, the required forcing is $F(t)=-2m\gamma\omega_0A\sin(\omega_0t-\phi)$.

		−	Conversely, if the forcing has amplitude $F_0$ and $\omega=\omega_0$, then $2m\gamma\omega A=F_0$, and the amplitude of the forced oscillation is $A=F_0/(2m\gamma\omega_0)$. When a constant force $F_0$ is applied instead, the deflection of the oscillator is $F_0/(m\omega_0^2)$. The amplitude of the forced oscillation is thus $\omega_0/(2\gamma)$ times this deflection.
		+	Conversely, if the forcing has amplitude $F_0$ and $\omega=\omega_0$, then $2m\gamma\omega_0A=F_0$, and the amplitude of the forced oscillation is $A=F_0/(2m\gamma\omega_0)$. When a constant force $F_0$ is applied instead, the deflection of the oscillator is $F_0/(m\omega_0^2)$. The amplitude of the forced oscillation is thus $\omega_0/(2\gamma)$ times this deflection.

			#### Answer

		−	[Insert a concise answer or boxed result]
		+	a. $F(t)=-2m\gamma\omega_0A\sin(\omega_0t-\phi)$
		+
		+	b. $A=F_0/(2m\gamma\omega_0)$, which is $\omega_0/(2\gamma)$ times the deflection under constant force

	@@ -1,13 +1,17 @@
	### Statement		### Statement

	$3.5.36.$ [Insert the problem statement]		$3.5.36.$ a. What is the necessary forcing force for an oscillator of mass $m$ with damping coefficient $\gamma$ to begin to perform harmonic oscillations with natural frequency $\omega_0$ according to the law $x=A\cos(\omega_0t-\phi)$?

			b. The amplitude of the forcing force is $F_0$, its frequency $\omega=\omega_0$. Determine the amplitude of the forced oscillations. How many times is it the deflection of the oscillator when a constant force $F_0$ is applied?

	### Solution		### Solution

	Substituting the particular (i.e. steady-state) solution $x=A\cos(\omega_0t-\phi)$ into the expression $\ddot x+2\gamma\dot x$+\omega_0^2x, we obtain $-2\gamma\omega A\sin(\omega_0t-\phi)$. (Note that the first and third terms cancel out, because the particular solution happens to be the solution to the homogeneous equation $\ddot x+\omega_0^2x=0$.) Thus, the required forcing is $F(t)=-2m\gamma\omega A\sin(\omega_0t-\phi)$.		Substituting the particular (i.e. steady-state) solution $x=A\cos(\omega_0t-\phi)$ into the expression $\ddot x+2\gamma\dot x+\omega_0^2x$, we obtain $-2\gamma\omega_0A\sin(\omega_0t-\phi)$. (Note that the first and third terms cancel out, because the particular solution happens to be the solution to the homogeneous equation $\ddot x+\omega_0^2x=0$.) Thus, the required forcing is $F(t)=-2m\gamma\omega_0A\sin(\omega_0t-\phi)$.

	Conversely, if the forcing has amplitude $F_0$ and $\omega=\omega_0$, then $2m\gamma\omega A=F_0$, and the amplitude of the forced oscillation is $A=F_0/(2m\gamma\omega_0)$. When a constant force $F_0$ is applied instead, the deflection of the oscillator is $F_0/(m\omega_0^2)$. The amplitude of the forced oscillation is thus $\omega_0/(2\gamma)$ times this deflection.		Conversely, if the forcing has amplitude $F_0$ and $\omega=\omega_0$, then $2m\gamma\omega_0A=F_0$, and the amplitude of the forced oscillation is $A=F_0/(2m\gamma\omega_0)$. When a constant force $F_0$ is applied instead, the deflection of the oscillator is $F_0/(m\omega_0^2)$. The amplitude of the forced oscillation is thus $\omega_0/(2\gamma)$ times this deflection.

	#### Answer		#### Answer

	[Insert a concise answer or boxed result]		a. $F(t)=-2m\gamma\omega_0A\sin(\omega_0t-\phi)$

			b. $A=F_0/(2m\gamma\omega_0)$, which is $\omega_0/(2\gamma)$ times the deflection under constant force