Respuesta :
Answer:
Angular velocity: approximately [tex]5.2360\; {\rm s^{-1}}[/tex].
Initial phase angle: [tex](\pi/2)[/tex].
Speed of the particle at [tex]x = 0.10\; {\rm m}[/tex]: approximately [tex]0.91\; {\rm m\cdot s^{-1}}[/tex].
Total energy: approximately [tex]0.16\; {\rm J}[/tex].
Explanation:
Assuming that this particle is in a simple harmonic motion, the position [tex]x[/tex] and velocity [tex]v[/tex] of this particle at time [tex]t[/tex] can be modelled as:
[tex]x(t) = A\, \cos(\omega\, t + \varphi)[/tex], and
[tex]v(t) = -A\, \omega\, \sin(\omega\, t + \varphi)[/tex],
Where:
- [tex]A[/tex] is the amplitude of the motion (maximum distance from equilibrium position,)
- [tex]\omega[/tex] is the angular frequency of the motion,
- [tex]\varphi[/tex] is the initial phase angle.
In this question, the amplitude of the motion would be [tex]A = 20\;{\rm cm} = 0.20\; {\rm m}[/tex].
If the period of a simple harmonic motion is [tex]T[/tex], the angular frequency of that motion would be [tex]\omega = (2\, \pi) / T[/tex]. In this question, given that the period of the motion is [tex]T = 1.2\; {\rm s}[/tex], the angular frequency of this motion would be:
[tex]\begin{aligned}\omega &= \frac{2\, \pi}{T} = \frac{2\, \pi}{1.2\; {\rm s}} \approx 5.2360\; {\rm s^{-1}}\end{aligned}[/tex].
It is given that position is [tex]0[/tex] at [tex]t = 0[/tex], meaning that [tex]\cos(\omega\, t + \varphi) = 0[/tex], which requires either [tex]\omega\, t + \varphi = (\pi/2)[/tex] or [tex]\omega\, t + \varphi = (3\, \pi / 2)[/tex]. Since velocity is positive, [tex]\sin(\omega \, t + \varphi) > 0[/tex]. Only [tex]\omega\, t + \varphi = (\pi/2)\![/tex] would meet this requirement.
Since [tex]t = 0[/tex], solve [tex]\omega\, t + \varphi = (\pi/2)[/tex] to obtain the initial phase angle [tex]\varphi[/tex]:
[tex]\displaystyle \varphi = \frac{\pi}{2} - \omega \, t = \frac{\pi}{2}[/tex].
In other words, the initial phase angle of this motion is [tex](\pi / 2)[/tex].
The speed of the particle at the given position [tex]x = 10\; {\rm cm} = 0.10\; {\rm m}[/tex] can be found from the expression for velocity, [tex]v(t) = -A\, \omega\, \sin(\omega\, t + \varphi)[/tex]. To do so, start by finding an expression for [tex](\omega\, t + \varphi)[/tex] from the equation for position [tex]x(t) = A\, \cos(\omega\, t + \varphi)[/tex] and substitute that value into the expression for [tex]v(t)[/tex].
Given that [tex]x = 0.10\; {\rm m}[/tex]:
[tex]A\, \cos(\omega\, t + \varphi) = x[/tex].
[tex]\displaystyle\omega\, t + \varphi = \arctan\left(\frac{x}{A}\right)[/tex].
Substitute this expression (as well as the expression [tex]\omega = (2\, \pi) / T[/tex]) into [tex]v(t) = -A\, \omega\, \sin(\omega\, t + \varphi)[/tex] to obtain:
[tex]\begin{aligned}v &= -A\, \omega\, \sin(\omega\, t + \varphi) \\ &= -A\, \left(\frac{2\, \pi}{T}\right)\, \sin\left(\arccos\left(\frac{x}{A}\right)\right) \\ &= -(0.20\; {\rm m}) \, \left(\frac{2\, \pi}{(1.2\; {\rm s^{-1}})}\right)\, \sin\left(\arccos\left(\frac{0.10}{0.20}\right)\right) \\ &\approx (-0.91)\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].
In other words, when [tex]x = 0.10\; {\rm m}[/tex], the velocity of the particle would be [tex](-0.91)\; {\rm m\cdot s^{-1}}[/tex]. The sign of velocity is negative because velocity represents a vector quantity, and that quantity points opposite to the positive direction. Since speed is the magnitude of velocity, the speed of the particle at that moment would be [tex]0.91\; {\rm m\cdot s^{-1}}[/tex] (without the negative sign since speed is a scalar quantity.)
While the spring constant of this oscillator system isn't given, the total energy of this system can be found from the kinetic energy of the oscillator. Specifically, because mechanical energy is conserved in simple harmonic motions, the total energy of the system would be equal to the maximum value of the kinetic energy achieved in the motion.
The kinetic energy of the oscillator is maximized when the oscillator is travelling at maximum speed [tex]A\, \omega[/tex]. The energy of the oscillator would be:
[tex]\begin{aligned} & (\text{total energy}) \\ =\; & (\text{maximum kinetic energy}) \\ =\; & \frac{1}{2}\, m\, v^{2} \\ =\; & \frac{1}{2}\, m\, \left(A\, \omega\right)^{2}\\ =\; & \frac{1}{2}\, m\, \left(\frac{2\, \pi\, A}{T}\right)^{2} \\ =\; & \frac{1}{2}\, (0.300\; {\rm kg}) \, \left(\frac{2\, \pi\, (0.20\; {\rm m})}{1.2\; {\rm s}}\right)^{2} \\ \approx\; & 0.16\; {\rm J}\end{aligned}[/tex].
