Respuesta :
Answer:
Explanation:
By using the Newton second law and the position equation for a simple harmonic motion we have
[tex]F=ma\\a_{max}=\omega^{2}A\\x=Acos(\omega t+ \phi)\\[/tex]
where a is the acceleration, w is the angular frecuency and \phi is the phase constant. We can calculate w from the equation for the maximum acceleration
[tex]\omega=\sqrt{\frac{a_{max}}{A}}=\sqrt{\frac{8*10^{3}m/s^{2}}{2*10^{-3}}}=2000rad/s[/tex]
(a).
[tex]F=ma=m\omega^{2}Acos(\omega t + \phi)\\F=(30*10^{-3}kg)(2*10^{-3}m)(2000\frac{rad}{s})^{2}cos(2000\frac{rad}{s} t - \frac{\pi}{2})=240N*cos(2000\frac{rad}{s} t - \frac{\pi}{2})[/tex]
(b). [tex]T=\frac{2\pi}{\omega}=\frac{2\pi}{2000}=3.14*10^{-13} s[/tex]
(c). [tex]v_{max}=A\omega=(2*10^{-3}m)(2000\frac{rad}{s})=4\frac{m}{s}[/tex]
(d). The mecanical energy is the kinetic energy when the velocity is a maximum
[tex]E_{m}=E_{k}(v_{max})=\frac{mv_{max}^{2}}{2}=\frac{30*10^{-3}kg(4\frac{m}{s})^{2}}{2}=0.024J[/tex]
Answer:
Explanation:
Given:
Mass, m = 30 g
= 0.03 kg
Amplitude, A = 2.0 ✕ 10^-3 m
Maximum acceleration, am = 8.0 ✕ 10^3 m/s2
Phase constant, phil = -π/2 rad
Displacement, x = A cos(wt + phil)
dx/dt = v = -Aw sin(wt + phil)
dv/dt = a = -Aw^2 cos(wt + phil)
Force, f = mass × acceleration
F = -0.03 × Aw^2 cos(wt + phil)
B.
At am,
a = Aw^2
8.0 ✕ 10^3 = 2.0 ✕ 10^-3 × w^2
w = sqrt(4 × 10^6)
= 2 × 10^3 rad/s
But w = 2pi/T
Where T = period
T = 2pi/2000
= 3.142 × 10^-3 s
= 0.00314 s
C.
Velocity = Aw
= 2.0 ✕ 10^-3 × 2000
= 4 m/s
D.
Total mechanical energy = kinetic energy + potential energy
= 1/2mv^2 + mgx
= 0.03 × [1/2 × 4^2 + (9.8 × (2 × 10^-3 × cos(2π - π/2)))]
= 0.03 × (8 + 0)
= 0.24 J
