Answer:
a) T = 1,467 s
, b) A = 0.495 m
, c) v = 4.97 10⁻² m / s
Explanation:
The simple harmonic movement is described by the expression
x = A cos (wt + Ф)
Where the angular velocity is
w = √ k / m
a) Ask the period
Angular velocity, frequency and period are related
w = 2π f = 2π / T
T = 2π / w
T = 2pi √ m / k
T = 2π √ (1.2 / 22)
T = 1,467 s
f = 1 / T
f = 0.68 Hz
b) ask the amplitude
The mechanical energy of a harmonic oscillator
E = ½ k A²
A = √2 E / k
A = √ (2 2.7 / 22)
A = 0.495 m
c) the mass changes to 8.0 kg
As released from rest Ф = 0, the equation remains
x = A cos wt
w = √ (22/8)
w = 1,658
x = 3.0 cos (1,658 t)
Speed is
v = dx / dt
v = -A w sin wt
The speed is maximum when without wt = ±1
v = Aw
v = 0.03 1,658
v = 4.97 10⁻² m / s
Measurement means weight, size, length, or capacity of something.
Answer:
M₀ = 5i - 4j - k
Explanation:
Using the cross product method, the moment vector(M₀) of a force (F) is about a given point is equal to cross product of the vector A from the point (r) to anywhere on the line of action of the force itself. i.e
M₀ = r x F
From the question,
r = i + j + k
F = 1i + 0j + 5k
Therefore,
M₀ = (i + j + k) x (1i + 0j + 5k)
M₀ = ![\left[\begin{array}{ccc}i&j&k\\1&1&1\\1&0&5\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Di%26j%26k%5C%5C1%261%261%5C%5C1%260%265%5Cend%7Barray%7D%5Cright%5D)
M₀ = i(5 - 0) -j(5 - 1) + k(0 - 1)
M₀ = i(5) - j(4) + k(-1)
M₀ = 5i - 4j - k
Therefore, the moment about the origin O of the force F is
M₀ = 5i - 4j - k