Respuesta :
Answer:
a) k = 120 N / m
, b) f = 0.851 Hz
, c) v = 1,069 m / s
, d) x = 0
, e) a = 5.71 m / s²
, f) x = 0.200 m
, g) Em = 2.4 J
, h) v = -1.01 m / s
Explanation:
a) Hooke's law is
F = k x
k = F / x
k = 24.0 / 0.200
k = 120 N / m
b) the angular velocity of the simple harmonic movement is
w = √ k / m
w = √ (120 / 4.2)
w = 5,345 rad / s
Angular velocity and frequency are related.
w = 2π f
f = w / 2π
f = 5.345 / 2π
f = 0.851 Hz
c) the equation that describes the movement is
x = A cos (wt + Ф)
As the body is released without initial velocity, Ф = 0
x = 0.2 cos wt
Speed is
v = dx / dt
v = -A w sin wt
The speed is maximum for sin wt = ±1
v = A w
v = 0.200 5.345
v = 1,069 m / s
d) when the function sin wt = -1 the function cos wt = 0, whereby the position for maximum speed is
x = A cos wt = 0
x = 0
e) the acceleration is
a = d²x / dt² = dv / dt
a = - Aw² cos wt
The acceleration is maximum when cos wt = ± 1
a = A w²
a = 0.2 5.345
a = 5.71 m / s²
f) the position for this acceleration is
x = A cos wt
x = A
x = 0.200 m
g) Mechanical energy is
Em = ½ k A²
Em = ½ 120 0.2²
Em = 2.4 J
h) the position is
x = 1/3 A
Let's calculate the time to reach this point
x = A cos wt
1/3 A = A cos 5.345t
t = 1 / w cos⁻¹(1/3)
The angles are in radians
t = 1.23 / 5,345
t = 0.2301 s
Speed is
v = -A w sin wt
v = -0.2 5.345 sin (5.345 0.2301)
v = -1.01 m / s
i) acceleration
a = -A w² sin wt
a = - 0.2 5.345² cos (5.345 0.2301)
a = -1.91 m / s²
The spring constant of this metal cylinder is equal to 120 N/m.
Given the following data:
Mass = 4.20 kg.
Distance = 0.200 m.
Force = 24.0 N.
How to calculate the spring constant.
Since the metal cylinder obeys Hooke’s law, its spring constant is given by this formula:
[tex]F=ke[/tex]
Where:
- k is the spring constant.
- x is the extension.
[tex]k=\frac{F}{e} \\\\k=\frac{24.0}{0.200}[/tex]
Spring constant = 120 N/m.
How to calculate the frequency of the oscillations.
First of all, we would determine the angular velocity of this metal cylinder:
[tex]\omega =\sqrt{\frac{k}{m} } \\\\\omega =\sqrt{\frac{120}{4.20} } \\\\\omega =5.345 \;rad / s[/tex]
For the frequency:
[tex]\omega = 2\pi f\\\\f=\frac{\omega}{2\pi} \\\\f=\frac{5.345}{2 \times 3.142}[/tex]
Frequency = 0.851 Hz.
How to calculate the maximum speed.
Mathematically, the maximum speed is given by this formula:
[tex]V_{max}=A\omega\\\\V_{max}=0.200 \times 5.345[/tex]
Maximum velocity = 1.069 m/s.
Note: [tex]cos \omega t = 0[/tex]
On the x-axis, the position at which the maximum speed would occur is zero (0).
How to calculate the maximum acceleration.
Mathematically, the maximum acceleration is given by this formula:
[tex]A_{max}=A\omega^2\\\\A_{max}=0.200 \times 5.345^2[/tex]
Maximum acceleration = 5.71 [tex]m/s^2[/tex].
The position for this maximum acceleration is given by:
[tex]x = A cos wt\\\\x = A[/tex]
x = 0.200 m.
How to calculate the total mechanical energy.
[tex]E=\frac{1}{2} kx^2\\\\E=\frac{1}{2} \times 120 \times 0.200^2[/tex]
E = 2.4 Joules.
How to calculate the speed of this cylinder.
[tex]x=\frac{1}{3} A[/tex]
First, we would determine the time:
[tex]x = A cos \omega t\\\\\frac{1}{3} A = A cos \omega t\\\\\frac{1}{3} = cos \omega t\\\\t=cos^{-1}( \frac{1}{3cos \omega} )\\\\t=cos^{-1}( \frac{1}{3cos 5.345} )[/tex]
Time, t = 0.2301 seconds.
For the speed:
[tex]V = -A w sin \omega t\\\\V = -0.2 5.345 sin (5.345\times 0.2301)[/tex]
Speed, V = -1.0 m/s.
How to calculate the acceleration of this cylinder.
[tex]x=\frac{1}{3} A[/tex]
[tex]a = -A \omega^2 sin \omegat\\\\a = - 0.2 \times 5.345² cos (5.345 \times0.2301)[/tex]
a = -1.91 [tex]m/s^2[/tex].
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