A system is represented by the equation:

2x”+128x = f(t)
Obtain the free response of the system in the form
a) x = A sin wn (t ) + B cos wn (t)
b) x= A sin wn (t) + ϕ
The initial conditions are x(0) = 0.05m and x’ = 0.3m/s.

a.

√(128/2) = 8.
x=(-0.3/8)sin8t+0.05cos8t
x=-0.0375sin8t+0.05cos8t

b.
A=+√( (0.05)^2 + ((-0.3)^2/(8)^2) ) = 0.062

sinϕ= (0.05/0.062) =0.806
cosϕ= (-0.3/0.062 *8) =-0.605

ϕtan= tan^-1 (0.806/-0.605) = tan^-1 (-1.333)=-0.927
-0.0927+π=2.217 rad

x= 0.062sin ( 8t + 2.214)m

Note: The values for part (a) are incorrect, as they do not satisfy the given initial conditions. Here is the corrected solution:

a.
The characteristic equation of the system is:
2λ^2 + 128 = 0
Solving for λ, we get:
λ = ±8i

Since the roots are imaginary, the general solution for the homogeneous equation is:
xh = c1 cos(8t) + c2 sin(8t)

To find the particular solution, we assume a sinusoidal form of the input:
f(t) = F sin(ωt)

Substituting into the equation of motion, we get:
-ω^2 F sin(ωt) + 128x = F sin(ωt)
x = (1/128-ω^2) F sin(ωt)

Comparing this to the form x = A sin(ωt) + B cos(ωt), we can see that:
A = 0
B = F/(128-ω^2)
ω = 8 (since we want the frequency of the output to be the same as the natural frequency of the system)

Therefore, the particular solution is:
xp = (1/1201) F sin(8t)

The total solution is the sum of the homogeneous and particular solutions:
x = xh + xp
x = c1 cos(8t) + c2 sin(8t) + (1/1201) F sin(8t)

Using the initial conditions x(0) = 0.05 and x' = 0.3, we get the following system of equations:
c1 + (1/1201) F = 0.05
8c2 = 0.3

Solving for c1 and c2, we get:
c1 = -0.0373
c2 = 0.0375

Therefore, the free response of the system is:
x = -0.0373 cos(8t) + 0.0375 sin(8t) + (1/1201) F sin(8t)

b.
Using the same particular solution as in part (a):
xp = (1/1201) F sin(8t)

We can write the total solution as:
x = A sin(8t + ϕ) + (1/1201) F sin(8t)

Using the initial conditions x(0) = 0.05 and x' = 0.3, we get the following system of equations:
A sin ϕ = 0.05
A ω cos ϕ + (1/1201) F = 0.3

Substituting ω = 8 and solving for A and ϕ, we get:
A = 0.062
ϕ = -0.925 (in radians) or -53.13°

Therefore, the free response of the system is:
x = 0.062 sin(8t - 0.925) + (1/1201) F sin(8t)

To obtain the free response of the system represented by the equation 2x'' + 128x = f(t), let's start by assuming that there is no external input, i.e., f(t) = 0. This will give us the homogeneous equation 2x'' + 128x = 0.

To find the solution for this differential equation, we can assume the solution has the form x = e^(st), where s is a constant to be determined. Substitute this assumed solution into the differential equation:

2(d^2x/dt^2) + 128x = 0

Taking the derivatives, we find:

2s^2e^(st) + 128e^(st) = 0

Dividing both sides by e^(st), we get:

2s^2 + 128 = 0

Solving this quadratic equation for s, we have:

s^2 = -64

Taking the square root of both sides, we get:

s = ±8i

Since s includes imaginary values, the general solution for x is of the form:

x = A e^(8it) + B e^(-8it)

Using Euler's formula e^(ix) = cos(x) + i sin(x), we can rewrite the general solution as:

x = A cos(8t) + i A sin(8t) + B cos(-8t) + i B sin(-8t)

Simplifying further using the identity cos(-x) = cos(x) and sin(-x) = -sin(x), we have:

x = A cos(8t) + i A sin(8t) + B cos(8t) - i B sin(8t)

Rearranging terms and combining like terms, we get:

x = (A + B) cos(8t) + i (A - B) sin(8t)

Since x(t) must be a real-valued function, the imaginary terms must cancel out. Therefore, we set A = B to get:

x = 2A cos(8t)

Now we need to use the initial conditions x(0) = 0.05 m and x' = 0.3 m/s to solve for A.

Plugging in t = 0 into the equation, we have:

0.05 = 2A cos(0)

Since cos(0) = 1, we can solve for A:

0.05 = 2A

A = 0.025

Thus, the free response of the system is given by:

x = 0.05 cos(8t) + 0.025 cos(8t)

Simplifying further:

x = 0.075 cos(8t)

Therefore, the free response of the system is represented by option a) x = A sin(wn(t)) + B cos(wn(t)), where A = 0 and B = 0.075.

Please note that option b) x = A sin(wn(t)) + ϕ is not applicable in this case as there is no phase angle specified in the given initial conditions.