Respuesta :
(a) 0.5 s
In a simple harmonic motion, the period is equal to the reciprocal of the frequency:
[tex]T=\frac{1}{f}[/tex]
where f is the frequency
For this simple harmonic oscillator, the frequency is
f = 2.0 Hz
So the period is
[tex]T=\frac{1}{2.0 Hz}=0.5 s[/tex]
(b) 12.56 rad/s
The angular frequency is given by
[tex]\omega = 2 \pi f[/tex]
where
f is the frequency
In this problem,
f = 2.0 Hz
So the angular frequency is
[tex]\omega = 2 \pi (2.0 Hz)=12.56 rad/s[/tex]
(d) 0.419 rad
The displacement of the system can be written as
[tex]x(t) = A cos (\omega t+\phi)[/tex] (1)
where A is the amplitude and [tex]\phi[/tex] is the phase constant.
The velocity is the derivative of the displacement:
[tex]v(t) = x'(t) = -A\omega sin(\omega t+\phi)[/tex] (2)
Here we know that
at t=0, x(5)=5.0 cm and v(t)=-28 cm/s. So we can rewrite the ratio (2)/(1) as
[tex]\frac{v(t)}{x(t)}=\frac{-28 cm/s}{5.0 cm}=\frac{-\omega A sin(\phi)}{A cos(\phi)}=-\omega tan \phi[/tex]
And re-arranging the equation we can find the phase constant:
[tex]\phi = tan^{-1} (\frac{v}{\omega x})=tan^{-1} (\frac{-28 cm/s}{(5.0 cm)(12.56 rad/s)})=0.419 rad[/tex]
(c) 5.47 cm
The displacement of the system can be written as
[tex]x(t) = A cos (\omega t+\phi)[/tex] (1)
at t=0, x=5.0 cm, so using the values we found for [tex]\omega, \phi[/tex] we can now solve the equation to find A, the amplitude:
[tex]A=\frac{x}{cos(\omega t+\phi)}=\frac{5.0 cm}{cos(0.419 rad)}=5.47 cm[/tex]
(e) 68.7 cm/s
The maximum speed in a simple harmonic system is given by
[tex]v=\omega A[/tex]
where in this case we have
[tex]\omega=12.56 rad/s[/tex]
[tex]A=5.47 cm[/tex]
Substituting the numbers into the formula, we find
[tex]v=(12.56 rad/s)(5.47 cm)=68.7 cm/s[/tex]
(f) 862.9 cm/s^2
The maximum acceleration in a simple harmonic system is given by
[tex]a=\omega^2 A[/tex]
where in this case we have
[tex]\omega=12.56 rad/s[/tex]
[tex]A=5.47 cm[/tex]
Substituting the numbers into the formula, we find
[tex]a=(12.56 rad/s)^2(5.47 cm)=862.9 cm/s^2[/tex]
(g) 0.04 J
The total energy of the system is equal to the kinetic energy when the speed of the system is maximum: this occurs at x=0 (equilibrium position), where the elastic potential energy is zero, and all the energy is just kinetic energy:
[tex]E=K=\frac{1}{2}mv_{max}^2[/tex]
where we have
m = 170 g = 0.170 kg is the mass
[tex]v_{max}=68.7 cm/s = 0.687 m/s[/tex] is the maximum speed
Substituting into the equation,
[tex]E=K=\frac{1}{2}(0.170 kg)(0.687 m/s)^2=0.04 J[/tex]
(h) 3.65 cm
The position of the system is given by
[tex]x(t) = A cos (\omega t+\phi)[/tex]
where we have
[tex]\omega=12.56 rad/s[/tex] is the angular frequency
[tex]A=5.47 cm[/tex] is the amplitude
[tex]\phi = 0.419 rad[/tex] is the phase constant
Substituting t=0.4 s, we find the position at this time:
[tex]x = (5.47 cm) cos ((12.56 rad/s)(0.4 s)+0.419 rad)=3.65 cm[/tex]
The SHM the oscillating and periodic motion of an object. a) T = 0.5 s. b) ω = 12.56 rad/seg. c) A = 5.476 cm. d) Φ = 0.4186 rad. e) Vmax = 68.778 cm/s. f) a = 863 m/s². g) Ek = 0.04012 J. h) X = 4.08 cm
What is simple harmonic motion, SHM?
It is a rectilinear movement performed by a oscillating and periodic movil.
It is periodic because it repeats in a certain intervale of time.
The time -in seconds- it takes to the movil to make a complete oscillation is called Period, T.
The frequency, f, refers to the number of complete oscillations (cycles) completed per second. f = 1/T (Hz).
When considering a simple harmonic motion, we need to know that there will be always a restoring force (F), that tends to take the movil to the original position when it moves a distance named amplitud.
There are some significant components to consider in this motion,
- The position of the movil, X (m) ⇒ initial, equilibrium, and in other positions
- The acceleration, a (m/s²)
- Velocity, V (m/s)
- The amplitud, A (m)
- Frequency, f (Hz)
- The initial phase, Φ (rad)
- Period, T (s)
Available data:
- m = 170 g
- f = 2 Hz
- t = 0 → x(t) = 5cm v(t) = -28 cm/s
(a) the period, T
T = 1/f
T = 1/2
T = 0.5 s
(b) the angular frequency, ω
ω = 2πf
ω = (2π)2Hz
ω = 12.56 rad/seg
(c) the amplitude, A
It can be cleared from the following displacement formula,
X = A sen (ω t + Φ)
First, we need to get Φ.
Φ = arctan (v/ωx) = tan⁻¹(v/ωx)
Φ = tan⁻¹(-28/12.56x5)
Φ = tan⁻¹(-28/62.8)
Φ = tan⁻¹(0.445)
Φ = 0.4186 rad
So now we can calculate A
X = A sen (ω t + Φ)
5 = A sen (12.56 x 0 + 0.4186)
5 = A sen (0.4186)
A = 5/cos(0.4186)
A = 5/0.913
A = 5.476 cm
(d) the phase constant, Φ
Φ = arctan (v/ωx) = tan⁻¹(v/ωx)
We did this in the previous step.
Φ = 0.4186 rad
(e) the maximum speed, Vmax
Vmax = A ω
Vmax = 5.476 x 12.56
Vmax = 68.778 cm/s
(f) the maximum acceleration
a = ω ² A
a = 12.56² x 5.476
a = 157.75 x 5.476
a = 863 m/s²
(g) the total energy
Total energy = kinetic energy + Potential energy = Ek + Ep
At x=0 ⇒ v=max, Ep = 0 ⇒ Et = Ek + Ep = Ek
Ek = 1/2 mv²
Ek = 1/2 (0.170 kg x 0.687²m/s)
Ek = 0.04012 J.
(h) the position at t = 0.4 s
X = A sen (ω t + Φ)
X = 5.476 cm sen (12.56 rad/seg x 0.4 + 0.4186 rad)
X = 5.476 sen (5.024 + 0.4186)
X = 5.476 sen (5.4426)
X = 4.08 cm
You can learn more about simple harmonic motion at
brainly.com/question/13858183
brainly.com/question/17315536
