Question

4. A spring is stretch 10 cm by a force of 3 N. A mass of 2 kg is hung from the spring and is also attached to a viscous damper that exerts a force of 3 N when the velocity of the mass is 5 m/s. If the mass is pulled down 5 cm below its equilibrium position and given an initial donward velocity of 10 cm/s determine its position u at any time t. Find the quasi-frequency μ and the ratio of μ to the natural frequency wo of the corresponding undamped motion. 5. A mass weighing 8 lb stretches a spring 1.5 in. The mass is also attached to a viscous damper with coefficient y. Determine the value of y for which the system is critically damped; be sure to give the (3.7.17) units for y

Answer 1

(4)

Given that,

L = 10 cm = 0.1 m

Fs = -3 N

m = 2 kg

Fd = -3 N

uo = 5 cm = 0.05 m

uo' = 10 cm/s = 0.1 m/s

ud = 5 m/s

Spring force, k = -Fs / L = 3 / 0.1 = 30 N/m

$\gamma$ = -Fd / ud = 3 / 5 = 0.6 Ns/m

Differential equation,

mx^2 + $\gamma$ x + k = 0

Put the values and multiply the euation with 5,

10x^2 + 3x + 150 = 0

By solving above quadratic equation,

x = -0.15 $\pm$ 3.87008

Position ,u(t) = Ae^-0.15t cos 3.87008t + Be^-0.15t sin3.87008t

Coefficient A = uo = 0.05 m

u(t) = 0.1 m/s = -0.15Ae⁰ cos0 - 3.87008Ae⁰ sin0 - 0.15Be⁰ cos0 + 3.87008 Be⁰ cos0

-0.15*0.05 = 3.87008B

B = 0.02777 m

Position at time t,

u(t) =  0.05 e^-0.15t cos 3.87008t + 0.0277 e^-0.15t sin3.87008t

u = Re^{- $\gamma$ t/2m} cos ( $\mu$ t - $\delta$ )

R = sqrt (A^2 + B^2) = 0.05719 m

$\delta$ = tan^-1 (B / A) = 0.50709

so, $\mu$ = sqrt (4km - $\gamma$ ^2) / 2m

$\mu$ = sqrt (4*30*2 - 0.6^2) / 2*2

$\mu$ = 3.87008 rad/s

Ratio of quasi frequency $\mu$ to wo,

$\mu$ / wo = 3.87008 / sqrt (k / m) =  3.87008 / sqrt (30 / 2)

$\mu$ / wo = 0.99925