A weight, oscillating freely on a spring, has moved from a distance of 0.5 cm from its equilibrium position to the largest one, equal to 1 cm, for a time of 0.01 s. What is the period of its oscillations?
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<h3> Statement </h3>
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$3.3.3$
A weight, oscillating freely on a spring, has moved from a distance of 0.5 cm from its equilibrium position to the largest one, equal to 1 cm, for a time of 0.01 s. What is the period of its oscillations?
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<h3>Solution</h3>
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For a harmonic motion, variable position depends sinusoidally on time. In this case, we suppose that load is at equilibrium position at $t = 0$. So, sine function is more adequate to this situation without phase angle or null phase angle. So,
$$x(t) = A\~\sin{\~\omega t}$$
where $A$ is the amplitude such that $A = 1\~{\rm{cm}}$, because it is the maximum distance. Let's suppose that $x_0 = 0.5\~{\rm{cm}}$ is achieved at $t=t_1$ and $A$ is achieved at $t = t_2$. Then,
$$x_0 = A\~\sin{\~\omega t_1} \\;(1)$$
$$A = A\~\sin{\~\omega t_2}$$ or
$$\sin{\~\omega t_2} = 1$$
Hence, $\omega t_2 = \frac{\pi}{2} + 2k\pi$ with $k\in\mathbb{Z}$, for $k=0$, $t_2 = \frac{\pi}{2\omega}$. As $\omega = \frac{2\pi}{T}$, so $t_2 = \frac{T}{4}$. Since $\Delta t = t_2 - t_1 = 0.01\~{\rm{s}}$,
$$t_1 = \frac{T}{4} - \Delta t \\;(2)$$
Putting (2) into (1) and separating $T$, it is obtained
$$T = \frac{4\Delta t}{1-\frac{2}{\pi}\arcsin{\frac{x_0}{A}}}$$
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<h4>Answer</h4>
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$$T = 0.06 {\rm{s}}$$
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<p style="text-align: right; font-style: italic; font-size: 14;">
BSc. Luis Daniel Fernández Quintana<br>
Physics Department (FCNE)<br>
Universidad de Oriente, Cuba<br>
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