Question

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=μ^2 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)

This question was addressed to me by my school teacher while I was bunking the class.

My enquiry is from Harmonic Motion and Mass in portion Linear Differential Equations – Second and Higher Order of Engineering Mathematics

Answer 1

Correct option is (b) y = e^(- λt) (y0 cos⁡vt+\(\left(\frac{y_1+ λy_0}{v}\right)\)  sin⁡vt)

To explain: 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=μ^2 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 m^2 + 2λm+ μ^2 = 0

\(m=\frac{-2λ±\sqrt{4λ^2-4μ^2}}{2} = – λ±vi\)…. given \(v = \sqrt{μ^2-λ^2}\)

yc = 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).