Правка разделов «Statement», «Solution», «Answer»

igor правка от 1 июн. 2026 г., 09:23

правка #18172 предыдущая #18170 ← раньше

en/3.3.31.md +41 −3

		@@ -1,11 +1,49 @@
			### Statement

		−	$3.3.31.$ [Insert the problem statement]
		+	$3.3.31.$ To measure small amplitudes of oscillations of a diaphragm making harmonic vibrations of high frequency $\omega$, a ”hammer” connected in an electrical circuit with a diaphragm and a telephone is used. The hammer of mass $m$ is pressed against the diaphragm with a force that is adjusted by a micrometer screw. When the contact of the hammer with the diaphragm is interrupted, the current in the circuit is interrupted and a rattling can be heard in the telephone. Determine the amplitude of the vibrations if the rattling begins when the force with which the hammer presses the diaphragm reaches the value $F$.

		+	![For problem $3.3.31$\|464x817, 50%](../../img/3.3.31/3.3.31.png)
		+
		+
			### Solution

		−	213
		+	This promlem is analogous to the previous one ([$3.3.30$](/en/3.3.30); if anything, look at the solution of the mentioned problem in russian). We can write for the hammer (projection onto the axis of oscillations)

		+	$$
		+	F-N=ma,
		+	$$
		+
		+	where $N$ is the normal reaction force of the diaphragm, $a$ is the acceleration of the hammer.
		+
		+	The diaphragm perform harmonic oscillations, thus
		+
		+	$$
		+	a=\omega^2A\sin\omega t,
		+	$$
		+
		+	where $A$ is the oscillation amplitude of the diaphragm, $t$ is time period.
		+
		+	From the previous equations we can obtain
		+
		+	$$
		+	N=F-m\omega^2A\sin\omega t.
		+	$$
		+
		+	If the value of $F$ is larger than $m\omega^2A$, then $N>0$ and there will be no the rattling (since $-1\leqslant\sin\omega t\leqslant1$). The rattling begins, when
		+
		+	$$
		+	F=m\omega^2A
		+	$$
		+
		+	and therefore
		+
		+	$$
		+	A=\frac{F}{m\omega^2}.
		+	$$
		+
		+
			#### Answer

		−	[Insert a concise answer or boxed result]
		+	$$
		+	A=\frac{F}{m\omega^2}.
		+	$$

	@@ -1,11 +1,49 @@
	### Statement		### Statement

	$3.3.31.$ [Insert the problem statement]		$3.3.31.$ To measure small amplitudes of oscillations of a diaphragm making harmonic vibrations of high frequency $\omega$, a ”hammer” connected in an electrical circuit with a diaphragm and a telephone is used. The hammer of mass $m$ is pressed against the diaphragm with a force that is adjusted by a micrometer screw. When the contact of the hammer with the diaphragm is interrupted, the current in the circuit is interrupted and a rattling can be heard in the telephone. Determine the amplitude of the vibrations if the rattling begins when the force with which the hammer presses the diaphragm reaches the value $F$.

			![For problem $3.3.31$\|464x817, 50%](../../img/3.3.31/3.3.31.png)


	### Solution		### Solution

	213		This promlem is analogous to the previous one ([$3.3.30$](/en/3.3.30); if anything, look at the solution of the mentioned problem in russian). We can write for the hammer (projection onto the axis of oscillations)

			$$
			F-N=ma,
			$$

			where $N$ is the normal reaction force of the diaphragm, $a$ is the acceleration of the hammer.

			The diaphragm perform harmonic oscillations, thus

			$$
			a=\omega^2A\sin\omega t,
			$$

			where $A$ is the oscillation amplitude of the diaphragm, $t$ is time period.

			From the previous equations we can obtain

			$$
			N=F-m\omega^2A\sin\omega t.
			$$

			If the value of $F$ is larger than $m\omega^2A$, then $N>0$ and there will be no the rattling (since $-1\leqslant\sin\omega t\leqslant1$). The rattling begins, when

			$$
			F=m\omega^2A
			$$

			and therefore

			$$
			A=\frac{F}{m\omega^2}.
			$$


	#### Answer		#### Answer

	[Insert a concise answer or boxed result]		$$
			A=\frac{F}{m\omega^2}.
			$$