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?

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

u=13

v=30

s/t=u+1\2at

=13+431/2

=13+265.5

=278.5

your answer is c! hope this : )