Solutions to Engineering Dynamics

ENGINEERING DYNAMICS 1
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2
3.4 Find the equation obtaining the force for the springs in parallel.
The springs in this case are parallel to each other hence, the total force is 2F because if
the spring is 1 it is F.
Let F = kx
In this case,2F = (Ke)x where Ke is the equivalent spring constant, hence, Ke = 2K
In parallel, spring constants get added.
Hence the equation for the force is given by F = kx
Hence, since F = kx , k = 10 and x = 4
Thus, F = 10 × 4
F = 40N
The oscillation period, T is given by:
Solving for T gives T = 2π √ m/k
Thus, 2 π √ 0.5/10 = 6.28 × 0.2236
=1.404s
3.15 Find the equation obtaining the force for the springs in series.
For springs in series, the total force on the mass is the same, F. However, the distance
movd is 2x, x due to each spring. Hence,
F= 2(ke)x
But for F = Kx to hold Ke = k/2 hence
1/ke = (1/k1) + (1/k2)
Hence the equation for the force is given by F = 2(ke)x

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The spring constants are 4cm and 12 cm respectively.
With a F= 40N, the total spring deformation is x1 + x2 = F (1/k1 + 1/k2)
= 40 (¼ + 1/ 12)
=13.33
The oscillation period, T is given by:
Solving for T gives T = 2π √ m/k
Thus, 2 π √ 1/10 = 6.28 × 0.31622
=1.986s
2.15 Determine the equations of particle in motion under the given conditions.
Mx = F(t) where m = 1.5 and F(t) = 12t – 24
1.5x = 12t – 24
X dv
dt = 6t – 12
V = ∫(6 t−12 dt )
V = 3t2 – 12t + c
When t = 1.5, the object is at rest, so v = 0.
0 = 12 -24 + c
C = 12
V = 3t2 – 12t + 12
X = ∫(3 t 2 – 12 t+12)dt
X = t3 – 6t2 +12t + c2
Initially when t= 0, x = 5.
Substituting gives c2 = 5