Question

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) \(\frac{7 sin t}{9}\)

The question was posed to me in an online interview.

I want to ask this question from Harmonic Motion and Mass in chapter Linear Differential Equations – Second and Higher Order of Engineering Mathematics

Answer 1

The correct answer is (b) \(\frac{7 sin t}{8}\)

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 λ^2=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 λ^2 = 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 \(x = \frac{21 sin t}{24} = \frac{7 sin t}{8}.\)