Question

A mass m = 1.2 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. At time t = 0 the mass is released from rest at a distance d = 0.35 m below its equilibrium height and undergoes simple harmonic motion with its position given as a function of time by y(t) = A cos(Ït â Ï). The positive y-axis points upward.a. Find the angular frequency of oscillations in radians per second.b. Determine the value of A in meters.c. Determine the value of Ï in radians.d. Enter an expression for velocity along y axis as function of time in terms of A, Ï and t using the value of Ï from part c.e. What is the velocity of mass at time t = 0.25 s?f. What is the magnitude of mass's maximum acceleration?

Answer 1

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

Answer 2

a. 3π/2 b. 0.36 m c. 0.234 m/s d. 42.55 m/s²

Step-by-step 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