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			<h3> Statement </h3>
			<p>
			$3.2.16.$ A heavy cart rolls with acceleration $a$ on an inclined plane forming an angle $\alpha$ with the horizon. Find the period of oscillation of a pendulum of length $l$ mounted on the cart.
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			For problem $3.2.16$
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			<h3>Solution</h3>
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			Acceleration vector $\vec{g}^*$ in an inertial frame of reference
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			<p>
			The period of oscillation of a pendulum on an inclined cart can be found using the well-known formula for the period of oscillation of a mathematical pendulum, by switching to the inertial frame of reference of the cart
			$$ T=2\pi\sqrt{\frac{l}{g^*}}\quad(1) $$
		@@ -83,7 +83,7 @@
			Where $\vec{g}^*$ is the acceleration of the inertial reference frame
			$$ \vec{g}^*=\vec{g}-\vec{a} $$
			Using the cosine law, we find the value of the modulus of the vector $\vec{g}^*$
		−	$$ (g^{*})^2=a^2+g^2-2ag\cos\alpha $$
		+	$$ (g^{*})^2=a^2+g^2-2ag\cos\alpha$$
			Where
			$$ \boxed{g^*=\sqrt{a^2+g^2-2ag\cos\alpha}} $$
			We substitute $(1)$ into the expression and find the period of oscillation of the pendulum
			$$ \boxed{T=2\pi\sqrt{\frac{l}{\sqrt{a^2+g^2-2ag\cos\alpha}}}} $$
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			<p style="text-align: right; font-style: italic; font-size: 14;">
			Dzikan Mikita<br>
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			<h4>Answer</h4>
			<p>
			$$T=2\pi\sqrt{l/{\sqrt{a^2+g^2-2ag\cos\alpha}}}$$
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	<h3> Statement </h3>		<h3> Statement </h3>
	<p>		<p>
	$3.2.16.$ A heavy cart rolls with acceleration $a$ on an inclined plane forming an angle $\alpha$ with the horizon. Find the period of oscillation of a pendulum of length $l$ mounted on the cart.		$3.2.16.$ A heavy cart rolls with acceleration $a$ on an inclined plane forming an angle $\alpha$ with the horizon. Find the period of oscillation of a pendulum of length $l$ mounted on the cart.
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	<figure>		<figure>
	<img src="https://savchenkosolutions.com/3/3.2.16/3.2.16.png"		<img src="https://savchenkosolutions.com/3/3.2.16/3.2.16.png"
	loading="lazy" width="180" />		loading="lazy" width="180" />
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	For problem $3.2.16$		For problem $3.2.16$
	</figcaption>		</figcaption>
	</figure>		</figure>
	</center>		</center>
	<p>		<p>
	</p>		</p>

	<h3>Solution</h3>		<h3>Solution</h3>
	<p>		<p>

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	<center>		<center>
	<figure>		<figure>
	<img src="https://savchenkosolutions.com/3/3.2.16/3.2.16_1.png"		<img src="https://savchenkosolutions.com/3/3.2.16/3.2.16_1.png"
	loading="lazy" width="210" />		loading="lazy" width="210" />
	<figcaption>		<figcaption>
	Acceleration vector $\vec{g}^*$ in an inertial frame of reference		Acceleration vector $\vec{g}^*$ in an inertial frame of reference
	</figcaption>		</figcaption>
	</figure>		</figure>
	</center>		</center>
	<p>		<p>
	The period of oscillation of a pendulum on an inclined cart can be found using the well-known formula for the period of oscillation of a mathematical pendulum, by switching to the inertial frame of reference of the cart		The period of oscillation of a pendulum on an inclined cart can be found using the well-known formula for the period of oscillation of a mathematical pendulum, by switching to the inertial frame of reference of the cart
	$$ T=2\pi\sqrt{\frac{l}{g^*}}\quad(1) $$		$$ T=2\pi\sqrt{\frac{l}{g^*}}\quad(1) $$
	@@ -83,7 +83,7 @@
	Where $\vec{g}^*$ is the acceleration of the inertial reference frame		Where $\vec{g}^*$ is the acceleration of the inertial reference frame
	$$ \vec{g}^*=\vec{g}-\vec{a} $$		$$ \vec{g}^*=\vec{g}-\vec{a} $$
	Using the cosine law, we find the value of the modulus of the vector $\vec{g}^*$		Using the cosine law, we find the value of the modulus of the vector $\vec{g}^*$
	$$ (g^{*})^2=a^2+g^2-2ag\cos\alpha $$		$$ (g^{*})^2=a^2+g^2-2ag\cos\alpha$$
	Where		Where
	$$ \boxed{g^*=\sqrt{a^2+g^2-2ag\cos\alpha}} $$		$$ \boxed{g^*=\sqrt{a^2+g^2-2ag\cos\alpha}} $$
	We substitute $(1)$ into the expression and find the period of oscillation of the pendulum		We substitute $(1)$ into the expression and find the period of oscillation of the pendulum
	$$ \boxed{T=2\pi\sqrt{\frac{l}{\sqrt{a^2+g^2-2ag\cos\alpha}}}} $$		$$ \boxed{T=2\pi\sqrt{\frac{l}{\sqrt{a^2+g^2-2ag\cos\alpha}}}} $$
	</p>		</p>
	<p style="text-align: right; font-style: italic; font-size: 14;">		<p style="text-align: right; font-style: italic; font-size: 14;">
	Dzikan Mikita<br>		Dzikan Mikita<br>
	</p>		</p>

	<h4>Answer</h4>		<h4>Answer</h4>
	<p>		<p>
	$$T=2\pi\sqrt{l/{\sqrt{a^2+g^2-2ag\cos\alpha}}}$$		$$T=2\pi\sqrt{l/{\sqrt{a^2+g^2-2ag\cos\alpha}}}$$
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