Respuesta :
1) Frequency: 1.95 Hz
2) Equilibrium position: 0.138 m
3) Amplitude: 0.357 m
4) Maximum speed: 4.36 m/s
5) Maximum acceleration: [tex]53.1 m/s^2[/tex]
6) Spring constant: 15.0 N/m
7) Total mechanical energy: 0.96 J
Explanation:
1)
The frequency of a simple harmonic motion is equal to the reciprocal of the period:
[tex]f=\frac{1}{T}[/tex]
where
f is the frequency
T is the period
For the particle in this problem, we have
T = 0.513 s (period)
So the frequency of motion is
[tex]f=\frac{1}{0.513}=1.95 Hz[/tex]
2)
In a simple harmonic motion, the object oscillates between two maximum positions [tex]+x_m[/tex] and [tex]-x_m[/tex] which are equidistance from the equilibrium position. So, the equilibrium position is the midpoint between these two positions.
For the particle in this problem, the two extreme positions are:
[tex]x_1 = -0.219 m[/tex]
[tex]x_2 = 0.495 m[/tex]
So the mid-point (the equilibrium position) is
[tex]x_{ep} = \frac{x_1 + x_2}{2}=\frac{-0.219+0.495}{2}=0.138 m[/tex]
3)
The amplitude of a simple harmonic motion is the maximum displacement of the object, measured from the equilibrium position.
This means that we can calculate the amplitude simply as the difference between one of the extreme positions and the equilibrium position.
Taking
[tex]x_2 = 0.495 m[/tex]
and
[tex]x_{ep} = 0.138 m[/tex]
We find the amplitude:
[tex]A=x_2 - x_{ep} = 0.495-0.138 =0.357 m[/tex]
4)
In a simple harmonic motion, the maximum speed is given by
[tex]v_{max}=\omega A[/tex]
where
[tex]\omega[/tex] is the angular frequency
A is the amplitude
The angular speed can be calculated from the frequency as follows:
[tex]\omega=2\pi f=2 \pi (1.95 Hz)=12.2 rad/s[/tex]
The amplitude is
A = 0.357 m
So, the maximum speed is
[tex]v_{max} = (12.2)(0.357)=4.36 m/s[/tex]
5)
The maximum acceleration in a simple harmonic motion is given by
[tex]a_{max}= \omega^2 A[/tex]
Where we already know that:
[tex]\omega=12.2 rad/s[/tex] is the angular frequency
A = 0.357 m is the amplitude of motion
Substituting, we find the maximum acceleration:
[tex]a_{max}=(12.2)^2(0.357)=53.1 m/s^2[/tex]
6)
The angular speed in a simple harmonic motion can be written as
[tex]\omega=\sqrt{\frac{k}{m}}[/tex]
where
k is the spring constant
m is the mass
Here we know that:
[tex]\omega=12.2 rad/s[/tex] is the angular speed
m = 0.101 kg is the mass of the particle
So we can solve the formula for k, the spring constant:
[tex]k=\omega^2 m =(12.2)^2(0.101)=15.0 N/m[/tex]
7)
Since the energy is conserved, the total mechanical mechanical energy is just equal to the maximum potential energy of the system, which occurs when the particle is at maximum displacement (x=A) and the speed is zero (so the kinetic energy is zero), therefore it is given by:
[tex]E=\frac{1}{2}kA^2[/tex]
where
k = 15.0 N/m is the spring constant
A = 0.357 m is the amplitude
And substituting,
[tex]E=\frac{1}{2}(15.0)(0.357)^2=0.96 J[/tex]
#LearnwithBrainly
