Правка раздела «Solution»
en/3.3.3.md
+2 −2
| ### Statement | |||
| $3.3.3$ A weight, oscillating freely on a spring, has moved from a distance of $0.5 \text{ cm}$ from its equilibrium position to the largest one, equal to $1 \text{ cm}$, for a time of $0.01 \text{ s}$. What is the period of its oscillations? | |||
| @@ -4,11 +4,11 @@Statement | |||
| ### Solution | |||
| − | 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, $$x_0 = A~\sin{~\omega t_1} | ||
| + | For a harmonic motion, variable position depends sinusoidally on time. In this case, we suppose that load is at equilibrium position at $t = 0$. So, sine function is more adequate to this situation without phase angle or null phase angle. So, $$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} \tag{1}$$ $$A = A~\sin{~\omega t_2} \Leftrightarrow \boxed{\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 | ||
| + | As $\omega = \frac{2\pi}{T}$, so $t_2 = \frac{T}{4}$. Since $\Delta t = t_2 - t_1 = 0.01~{\rm{s}}$, $$t_1 = \frac{T}{4} - \Delta t \tag{2}$$ | ||
| Putting $(2)$ into $(1)$ and separating $T$, it is obtained $$T = \frac{4\Delta t}{1-\frac{2}{\pi}\arcsin {\frac{x_0}{A}}}$$ | |||
| #### Answer | |||
| $$T = 0.06 \text{ s}$$ | |||
| ### Statement | ### Statement | ||
| $3.3.3$ A weight, oscillating freely on a spring, has moved from a distance of $0.5 \text{ cm}$ from its equilibrium position to the largest one, equal to $1 \text{ cm}$, for a time of $0.01 \text{ s}$. What is the period of its oscillations? | $3.3.3$ A weight, oscillating freely on a spring, has moved from a distance of $0.5 \text{ cm}$ from its equilibrium position to the largest one, equal to $1 \text{ cm}$, for a time of $0.01 \text{ s}$. What is the period of its oscillations? | ||
| @@ -4,11 +4,11 @@Statement | |||
| ### Solution | ### Solution | ||
| 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, $$x_0 = A~\sin{~\omega t_1} |
For a harmonic motion, variable position depends sinusoidally on time. In this case, we suppose that load is at equilibrium position at $t = 0$. So, sine function is more adequate to this situation without phase angle or null phase angle. So, $$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} \tag{1}$$ $$A = A~\sin{~\omega t_2} \Leftrightarrow \boxed{\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 |
As $\omega = \frac{2\pi}{T}$, so $t_2 = \frac{T}{4}$. Since $\Delta t = t_2 - t_1 = 0.01~{\rm{s}}$, $$t_1 = \frac{T}{4} - \Delta t \tag{2}$$ | ||
| Putting $(2)$ into $(1)$ and separating $T$, it is obtained $$T = \frac{4\Delta t}{1-\frac{2}{\pi}\arcsin {\frac{x_0}{A}}}$$ | Putting $(2)$ into $(1)$ and separating $T$, it is obtained $$T = \frac{4\Delta t}{1-\frac{2}{\pi}\arcsin {\frac{x_0}{A}}}$$ | ||
| #### Answer | #### Answer | ||
| $$T = 0.06 \text{ s}$$ | $$T = 0.06 \text{ s}$$ | ||