Harmonic Motion & Mass - Ordinary Differential Equations Multiple Choice Questions

This set of Ordinary Differential Equations MCQs focuses on “Harmonic Motion and Mass – Spring System”.

1. A particle moves along the x-axis according to the law \(\frac{d^2 x}{dt^2} + 6 \frac{dx}{dt} + 25x = 0.\) If the particle is started at x=0 with an initial velocity of 12ft/sec to the left, determine x in terms of t.
a) x=-3e^-3t sin⁡4t
b) x=-e^-3t (sin⁡4t+cos 4t)
c) x=-3e^-3t (sin⁡4t+cos⁡4t)
d) x=-3e^-3t cos⁡4t
View Answer

Answer: a
Explanation: x” (t) + 6x’ (t) + 25x(t) = 0 and at t=0, \(\frac{dx}{dt}\) = -12….since it is moving to the left.
m²+6m+25=0…..A.E
\(m=\frac{-6±\sqrt{6^2-4*25}}{2} = \frac{-6±8i}{2} = -3±4i\)
complementary solution x_c = e^-3t (c₁ cos⁡4t + c₂ sin⁡4t)
at t=0, x=0 therefore c₁ = 0 –> x_c = e^-3t c₂ sin⁡4t
differentiating \(\frac{dx_c}{dt}\) = -3e^-3t c₂ sin⁡4t + 4e^-3t c₂ cos⁡4t
at t=0 \(\frac{dx_c}{dt}\) = -12… given R.H.S = 4c₂ –> c₂ = -3
x_c = x = -3e^-3t sin⁡4t……as there is no particular solution.

2. Solve the problem of resonance damped vibration of a spring .If the governing D.E is given by \(m \frac{d^2 y}{dt^2} + c \frac{dy}{dt} + ky=0;\) c>0 with initial conditions as y(0)=y₀ and y’(0)=y₁ and assume c/m=2λ, k/m=μ² and \(v = \sqrt{μ^2-λ^2}\).
a) y = e^-λt (y₀ cos⁡vt + y₁ sin⁡vt)
b) y = e^(- λt) (y₀ cos⁡vt+\(\left(\frac{y_1+ λy_0}{v}\right)\) sin⁡vt)
c) y = e^(- λt) (\(\left(\frac{y_1+ λy_0}{v}\right)\) cos⁡vt + y₀ sin⁡vt)
d) y = e^-λt (y₀ cos⁡vt+λy₁ sin⁡vt)
View Answer

Answer: b
Explanation: The governing D.E is given by \(m \frac{d^2 y}{dt^2} + c \frac{dy}{dt} + ky=0;\) c>0
–> \( \frac{d^2 y}{dt^2} + \frac{c}{m} \frac{dy}{dt} + \frac{k}{m}y = 0\) given c/m=2λ, k/m=μ² thus D.E takes the form
\( \frac{d^2 y}{dt^2} + 2λ d\frac{dy}{dt} + μ^2 y = 0\) it’s A.E is m² + 2λm+ μ² = 0
\(m=\frac{-2λ±\sqrt{4λ^2-4μ^2}}{2} = – λ±vi\)…. given \(v = \sqrt{μ^2-λ^2}\)
y_c = y = e^{– λt} (c₁ cos⁡vt + c₂ sin⁡vt ) to find constants we use initial condition
at t=0 y = y₀ –> c₁ = y₀ therefore y = e^-λt (y₀ cos⁡vt + c₂ sin⁡vt)
now differentiate and substituting y’(0)=y₁
y’ = – λe^{– λt} (y₀ cos⁡vt + c₂ sin⁡vt) + e^{– λt} (-y₀ vsin⁡vt + vc₂ cos⁡vt)
at t=0, y₁ = – λy₀ + vc₂ –> c₂ = \(\left(\frac{y_1+ λy_0}{v}\right)\) thus y=e^{– λt} (y₀ cos⁡vt+\(\left(\frac{y_1+ λy_0}{v}\right)\) sin⁡vt).

3. Solve the problem of un-damped forced vibrations of a spring in the case where the forcing function is f(t)=A sin ωt. D.E associated with the problem is \(m \frac{d^2 y}{dt^2} + ky = f(t)\), with initial conditions as y(0)=y₀ and y’(0)=y₁ and assume λ² = k/m, μ=A/m.
a) y = y₀ cos⁡λt + y₁ sin⁡λt + \(\frac{λ cos ωt}{-ω^2+λ^2}\)
b) y = y₀ cos⁡λt + (y₁/μ)sin⁡λt + \(\frac{cos ωt}{ω^2+λ^2}\)
c) y = y₀ cos⁡λt + (y₁ λ)sin⁡λt +\(\frac{sin ωt}{ω^2+λ^2}\)
d) y = y₀ cos⁡λt + (y₁/λ)sin⁡λt + \(\frac{μ sin ωt}{-ω^2+λ^2}\)
View Answer

Answer: d
Explanation: \(m \frac{d^2 y}{dt^2} + ky = A \,sin \,ωt\, \,or\, \frac{d^2 y}{dt^2} + \frac{k}{m} y = \frac{A}{m} sin ωt \rightarrow \frac{d^2 y}{dt^2} + λ^2 y = μ sin ωt\)
A.E is m² + λ² = 0 –> m = ±λi –> y = c₁ cos⁡λt + c₂ sin⁡λt at t=0 y=y₀
–> c₁ = y₀ therefore y_c = y₀ cos⁡λt + c₂ sin⁡λt
now differentiate and substituting y’(0)=y₁
y’ = -y₀ λsin⁡λt + λc₂ cos⁡λt at t = 0 y₁/λ = c₂ and y_c = y₀ cos⁡λt + (y₁/λ)sin⁡λt
to find particular solution \(y_p = \frac{μ sin ωt}{D^2+λ^2}\) assuming μ as constant \(y_p = \frac{μ sin ωt}{-ω^2+λ^2}\)
∴y=y₀ cos⁡λt + (y₁/λ)sin⁡λt + \(\frac{μ sin ωt}{-ω^2+λ^2}\).

4. A particle undergoes forced vibrations according to the law x”(t) + 25x(t) = 21 sin⁡t. If the particle starts from rest at t=0, find the displacement at any time t>0.
a) \(\frac{21 cos t}{25}\)
b) \(\frac{7 sin t}{8}\)
c) \(\frac{3 cos t}{8}\)
d) \(\frac{7 sin t}{9}\)
View Answer

Answer: b
Explanation: x” (t) + 25x(t) = 21 cos⁡t comparing it with \( \frac{d^2 y}{dt^2} + \frac{k}{m} y= \frac{A}{m} sin ωt\)
\( \frac{d^2 y}{dt^2} + λ^2 y = μ sin \,ωt\) where λ²=k/m, μ=A/m and its solution is
y=y₀ cos⁡λt + (y₁/λ)sin⁡λt + \( \frac{μ sin ωt}{-ω^2+λ^2}\) where y(0)=y₀ and y’(0)=y₁.
we got λ² = 25 –> λ=5 and μ=21, ω=1. Its solution is
x = x₀ cos⁡5t + (x₁/5)sin⁡5t + \(\frac{21 sin t}{-1^2+25}\) particle is at rest at t = 0 –> x(0) = 0 = x₀
and x’(0) = 0 = x₁ therefore \(x = \frac{21 sin t}{24} = \frac{7 sin t}{8}.\)

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