Question

A single-degree freedom spring-mass system is subjected to a sinusoidal force of 10 N amplitude and frequency ω along the axis of the spring. The stiffness of the spring is 150 N/m, damping factor is 0.2 and the undamped natural frequency is10ω At steady state, the amplitude of vibration (in m) is approximately
1. 0.05
2. 0.07
3. 0.70
4. 0.90

Answer 1

Correct Answer - Option 2 : 0.07

Concept:

The amplitude ‘A’ of steady-state vibration in harmonic excitation with damper is,

\(A = {x_{max}} = \frac{{{F_0}}}{{\sqrt {{{\left( {K - m{\omega ^2}} \right)}^2} + {{\left( {C\omega } \right)}^2}} }}\)

Or, \(A = \frac{{{F_0}/K}}{{\sqrt {{{\left( {1 - {{\left( {\frac{\omega }{{{\omega _n}}}} \right)}^2}} \right)}^2} + {{\left( {2\xi \frac{\omega }{{{\omega _n}}}} \right)}^2}} }}\)

also, \(A = \frac{\delta }{{\sqrt {{{\left( {1 - {r^2}} \right)}^2} + {{\left( {2\xi r} \right)}^2}} }}\)

Where δ is static deflection of spring in m, F₀ is sinusoidal force amplitude in N.

, ω is forced frequency and ω_n is natural frequency of spring in rad/s

And ξ is damping factor = C/C_C = actual damping coefficient/critical damping coefficient.

Calculation:

Given, force amplitude F₀ = 10 N, K = 150 N/m, ξ = 0.2 and ω_n = 10 ω

From, \(A = \frac{{{F_0}/K}}{{\sqrt {{{\left( {1 - {{\left( {\frac{\omega }{{{\omega _n}}}} \right)}^2}} \right)}^2} + {{\left( {2\xi \frac{\omega }{{{\omega _n}}}} \right)}^2}} }}\)

\(A = \frac{{10/150}}{{\sqrt {{{\left( {1 - {{\left( {\frac{{\rm{\omega }}}{{10{\rm{\omega }}}}} \right)}^2}} \right)}^2} + {{\left( {2 \times 0.2 \times \frac{{\rm{\omega }}}{{10{\rm{\omega }}}}} \right)}^2}} }}\)

\(A = 0.0673 \approx 0.07 m\)