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.

Savchenko Solutions

Solutions of Savchenko Problems in Physics
knowledge must be free

    <h3 id="back-link"><a href="/#3.2">$\leftarrow$Back</a></h3>

    <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.

    <h3>Solution</h3>
    <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 $$T=2\pi\sqrt{\frac{l}{g^*}}\quad(1)$$ 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 $\vec{g}^*$ $$(g^{*})^2=a^2+g^2-2agcos\alpha$$ Where $$\boxed{g^*=\sqrt{a^2+g^2-2agcos\alpha}}$$ We substitute $(1)$ into the expression and find the period of oscillation of the pendulum $$\boxed{T=2\pi\sqrt{\frac{l}{\sqrt{a^2+g^2-2agcos\alpha}}}}$$

Dzikan Mikita

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


<footer class="row container">
  <br>
    <p>
        <small> © <strong>Savchenko Solutions</strong>, 2023-2024 <br></small>
    </p>
    <p>
        <small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> alex@savchenkosolutions.com <br></small>
    </p>
</footer>