Respuesta :
Answer:
a. 3π/2 b. 0.36 m c. 0.234 m/s d. 42.55 m/s²
Explanation:
Here is the complete question
A mass m = 1.1 kg hangs at the end of a vertical spring whose top end is fixed to the ceiling. The spring has spring constant k = 130 N/m and negligible mass. The mass undergoes simple harmonic motion when placed in vertical motion, with its position given as a function of time by y ( t ) = A c o s ( ω t − ϕ ) , with the positive y-axis pointing upward. At time t = 0 the mass is observed to be passing through its equilibrium height with an upward speed of v 0 = 3.9 m/s.
A. Find the smallest positive value of ϕ , in radians.
B. Calculate the value of A in meters.
C. What is the mass's velocity along the y-axis in meters per second, at the time t = 0.15 s?
D. What is the magnitude of the mass's maximum acceleration, in meters per second squared?
Solution
a. Since y ( t ) = A c o s ( ω t − ϕ ), the smallest possible value for ϕ is gotten when c o s ( ω t − ϕ ) = 0 ⇒ ω t − ϕ = cos⁻¹ 0 = π/2
ω t − ϕ = π/2.
At t = 0, ω t − ϕ = ω 0 − ϕ = 0 − ϕ = π/2
− ϕ = π/2
ϕ = -'π/2
Since this is a negative angle, we add 2π to the right side.
So, ϕ = -'π/2 + 2π = 3π/2
ϕ = 3π/2
b. Since v = Aω = A√(k/m) where v = maximum velocity at time t = 0 = 3.9 m/s. A = amplitude, k = spring constant = 130 N/m and m = mass = 1.1 kg
A = v/√(k/m) = 3.9 m/s/√(130 N/m/1.1 kg) = 3.9/√118.18 = 3.9/10.87 = 0.36 m
c. To find its velocity, we differentiate y(t)
So, v = dy(t)/dt = dA c o s ( ω t − ϕ )/dt = -'ωAsin( ω t − ϕ ) = v₀sin( ω t − ϕ )
v = v₀sin( ω t − ϕ ) = v₀sin( ω t − ϕ)
Substituting the value of the variables,
v = 3.9sin( 10.87 t − 3π/2)
At t = 0.15 s,
v = 3.9sin( 10.87 × 0.15 − 3π/2)
v = 3.9sin( 1.6305 − 4.7124)
v = -'3.9sin( -3.0819)
v = -'3.9 × - 0.06
v = 0.234 m/s
d. The maximum acceleration, a
a = Aω² = Ak/m = 0.36 × 130/1.1 = 42.55 m/s²
Answer:
a) F = 10.4 rad/s
b) A = 0.375 m
c) ϕ = 3π/2
d) V(t) = -ωAsin( ω t - 3π/2 )
e) V = 0.144 m/s
f) a = 40.625 m/s²
Explanation: Given that
mass m = 1.2 kg
The spring constant k = 130 N/m Time t = 0
Distance d = 0.35 m
y( t ) = A c o s ( ω t − ϕ )
At time t = 0
Speed of Vo = 3.9 m/s.
a) Find the angular frequency of oscillations in radians per second
W = √(k/m)
2πF = √(k/m)
F = 1/2π√(k/m)
F = 1/2π √(130/1.2)
F = 1.66Hz
1 Henz = 6.28 rad/s therefore,
F = 1.66 × 6.28
F = 10.4 rad/s
b) Calculate the value of A in meters.
V = Aω = A√(k/m)
where V = 3.9 m/s the maximum velocity at time t = 0
A = amplitude
A = v/√(k/m)
A = 3.9/√(130/1.2)
A = 3.9/10.4
A = 0.375 m
c. Determine the value of ϕ in radians
If y( t ) = A c o s ( ω t − ϕ ) We can obtain the smallest possible value of ϕ when c o s ( ω t − ϕ ) = 0
ω t − ϕ = cos⁻¹ 0 = π/2
ω t − ϕ = π/2.
At t = 0,
ω(0) − ϕ = π/2
− ϕ = π/2
ϕ = -'π/2
This is a negative angle, let us add 2π to the right side. So,
ϕ = -'π/2 + 2π = 3π/2
ϕ = 3π/2
d. Enter an expression for velocity along y axis as function of time in terms of A, ϕ and t using the value of ϕ from part c.
To find expression for velocity, we differentiate y(t) with respect to time t So,
V = dy/dt = dA c o s ( ω t − ϕ )/dt
V = -ωAsin( ω t − ϕ )
Therefore
V(t) = -ωAsin( ω t + π/2 )
Or
V(t) = -ωAsin( ω t - 3π/2 ) ...... (1)
e. What is the velocity of mass at time t = 0.25 s?
From equation (1)
V(t) = V₀sin( ω t − ϕ )
Substituting the value of the variables,
V = 3.9sin( 10.4t − 3π/2)
At t = 0.25 s,
V = 3.9sin( 10.4 × 0.25 − 3π/2)
V = 3.9sin( 2.6 − 4.7124)
V= -3.9sin( -2.1124)
V= -3.9 × - 0.037
V = 0.144 m/s
f. What is the magnitude of mass's maximum acceleration?
The maximum acceleration a = Aω²
a = Aω² =
a = Ak/m
a = 0.375 × 130/1.2
a = 40.625 m/s²
