(b) What is the velocity of the block (in m/s) at t = 2.00 s?(c) What is the acceleration of the block (in m/s2) at t = 2.00 s? A simple harmonic oscillator consists of a spring with a force constant of k = 59.5 N/mconnected to a block with a mass of m = 1.05 kg. Assume that the surface supporting theblock is flat and frictionless, and there is no air resistance. At time t = 0, the spring isstretched 16.5 cm from its natural length and the block is moving at a speed of 1.95 m/sin the -x direction, as shown below.1.95 m/s+16.5 cm(a) Find x(t), the displacement of the block from equilibrium as a function oftime. Hint: you’ll need to find the constants o (in rad/s), A (in cm), ø (inradians) for the function: x(t) = Acos(@t + p).

Let the mass of the block be m. Let the displacement of the block from the equilibrium position be x and the spring constant is k. The differential equation for the motion of the block is md2xdt2+kx=0 This gives us d2xdt2+kmx=0⇒d2xdt2+ω2x=0 ω is the natural frequency of oscillation. The solution of this differential equation is of the form xt=A cosωt+ϕ A is known as the amplitude of the oscillation and ϕ is the phase angle. These are undetermined constants and can be found using appropriate boundary conditions.a. In the given question, we have the mass of the block m=1.05 kg the spring constant of the spring k=59.5 N/m Therefore the frequency of oscillation ω=km=59.51.05s-1=7.527 s-1 We have the displacement as a function of time xt=A cosωt+ϕ We have two boundary conditions x0=16.5 cm=16.5100m=0.165 mx˙0=v0=dxdtt=0=1.95 m/s This gives us A cosϕ=0.165-ωA sinϕ=-7.527Asinϕ=1.95⇒Asinϕ=-1.957.527=-0.259 Taking the ratio between the two we have tanϕ=-0.2590.165=-1.56⇒ϕ=tan-1-1.56=-57.5° In radians, this is equal to ϕ=π180×57.5=-1.004 rad Therefore we can find the amplitude using Acos-57.5°=0.165⇒A=0.165cos-57.5°=0.307 m Thus the displacement as a function of time is xt=0.307 cos7.527t-1.004 b. To find the velocity of the block we differentiate this expression once dxdt=v(t)=-0.307×7.527×sin7.527t-1.004=-2.311 sin7.527t-1.004 Therefore at t=2 s the velocity is v2=-2.311 sin7.527×2-1.004=-2.311 sin14.05=-2.302 m/s c. Acceleration is given by a=dvdt=-2.311×7.527×cos7.527t-1.004=-17.395 cos7.527t-1.004 Therefore at t=2 s the acceleration is a2=-17.395 cos7.527×2-1.004=-17.395 cos14.05=-1.516 m/s2 Answer: a. The displacement as a function of time is xt=0.307 cos7.527t-1.004. b. Velocity of the block after 2 seconds is v2=-2.302 m/s. c. The acceleration of the block after 2 seconds is a2=-1.516 m/s2.

Answered: (a) Find x(t), the displacement of the…

(a) Find x(t), the displacement of the block from equilibrium as a function of time. Hint: you’ll need to find the constants o (in rad/s), A (in cm), ø (in radians) for the function: x(t) = Acos(@t + 4). %3D