Question

A single degree of freedom system having mass 1 kg and stiffness $10\frac{kN}{m}$ initially at rest is subject to an impulse force of magnitude 5kN for ${10}^{-4}$ seconds. The amplitude in mm of the resulting free vibration is

• A. 0.5
• B. 1.0
• C. 5.0
• D. 10.0

Answer 1

The correct option is C 5.0


At t = 0 mass is at rest F = 5 kN for ${10}^{-4}$ seconds

Amplitude = x

$F=m\times \frac{dv}{dt}$

$F.{\int }_{t_1}^{t_2}dt=m.{\int }_{v_1}^{v_2}dv$

$F(t_2-t_1)=m\times (v_2-v_1)$

$5\times {10}^3\times {10}^{-4}=1\times (v_2-0)$

$0.5m/s=v_2$

$\frac{1}{2}m{v}_2^2=\frac{1}{2}k{x}^2$

$x=\sqrt{\frac{m{v}_2^2}{k}}=\sqrt{\frac{1\times 0.5\times 0.5}{10\times 1000}}$

$=\frac{0.5}{100}=0.005m=5mm$