Respuesta :
A 0.293-kg particle undergoes simple harmonic motion along the horizontal x-axis between the points x1 = -0.285 m and x2 = 0.395 m. The period of oscillation is 0.641 s. Find the frequency, f, the equilibrium position, xeq, the amplitude, A, the maximum speed, vmax, the maximum magnitude of acceleration, amax, the force constant, k, and the total mechanical energy, Etot.
That being said,
f=1/T=1/0.641=1.56 s
xeq=(x1+x2)/2=(-0.285+0.395)/2=0.055 m
A=(x2-x1)/2=(0.395+0.285)/2=0.340 m
x(t)=Acos(wt+ϕ)
v(t)=-wAsen(wt+ϕ)
w=2πf=2*3.14*1.56=9.8 rad/s
vmax=-wA=9.8*0.340=-3.3 m/s
a(t)=-w^2Acos(wt+ϕ)
amax=-w^2A=96.04*0.340=-32.6 m/s^2
w=(k/m)^1/2 => k=mw^2=0.293*96.04=28.1 N/m
Etot=1/2kA^2
Etot=0.5*28.1*0.116=1.63 J
The characteristics of the simple harmonic movement allow to find the results for the questions about the movement of the mass with the spring are:
a) The frequency is: f = 1.72 Hz
b) The equilibrium position is: x₀ = 0.088 m
c) The amplitude is: A = 0.686 m
d) The maximum speed is: v = 7.42 m / s
e) the maximum acceleration is: a = 80.16 m / s²
f) The mechanical energy is: E = 3.44 J
Given parameters
- Particle mass m = 0.125 kg
- Initial point x₀ = -0.299 m
- End point x_f = 0.387 m.
- Oscillation period T = 0.581 s
To find.
a) Frequency.
b) the equilibrium position.
c) The amplitude.
d) Maximum speed.
e) Maximum acceleration.
f) The total mechanical energy.
Simple harmonic motion is a periodic motion where the restoring force is proportional to the elongation. This movement is fully described by the expression
x = A cos (wt + Ф)
w² = [tex]\sqrt{ \frac{k}{m} }[/tex]
Where x is the displacement, A the amplitude of the movement, w the angular velocity, t the time, Ф a phase constant that meets the initial conditions, k the spring constant and m the mass attached to the spring.
a) Frequency and period are related.
[tex]f= \frac{1}{T}[/tex]
Let's calculate.
[tex]f = \frac{1}{0.581 }[/tex]
f = 1.72 Hz
b) The equilibrium position occurs at the midpoint of the movement
[tex]x_o = \frac{x_f + x_o}{2}[/tex]
[tex]x_o = \frac{0.387 + (-0.299)}{2}[/tex]
x₀ = 0.088 m
c) The amplitude is the maximum elongation of the spring
[tex]A= x_f - x_o[/tex]
A = 0.387 - (-0.299)
A = 0.686 m
d) The speed is defined by the variation of the position with respect to time.
[tex]v = \frac{dx}{dt}[/tex]
Let's make the derivatives
v = - A w cos (wt + Ф)
The angular velocity is related to the period.
[tex]w= \frac{\pi }{T} \\w= \frac{2 \pi }{0.581}[/tex]
w = 10.81 rad / s
The speed is maximum when the cosine is maximum ±1
v = A w
v = 0.686 10.81
v = 7.42 m / s
e) Acceleration is defined as the change in velocity over time.
[tex]a = \frac{dv}{dt}[/tex]
a = -A w² sin (wt + Ф)
The acceleration is maximum when the sine function is maximum ±1
a = A w²
a = 0.686 10.81²
a = 80.16 m / s²
f) The mechanical energy is calculate for the point of maximum elongation.
E = ½ k A²
k = w² m
Let's replace.
E = ½ w² m A²
Let's calculate.
E = ½ 10.81² 0.125 0.686²
E = 3.44 J
In conclusion, using the characteristics of simple harmonic motion we can find the results for the questions about the motion of the mass with the spring are:
a) The frequency is: f = 1.72 Hz
b) The equilibrium position is: x₀ = 0.088 m
c) The amplitude is: A = 0.686 m
d) The maximum speed is: v = 7.42 m / s
e) the maximum acceleration is: a = 80.16 m / s²
f) The mechanical energy is: E = 3.44 J
Learn more about simple harmonic motion here: brainly.com/question/17315536
