Question 2.1.2: Consider the forced vibration of a mass m connected to a spr......

From equation (2.11) the response with x_{0} = ν_{0} = 0 becomes

x(t) = \frac{ν_{0}}{w_{n}}\sin w_{n}t  +  \left(x_{0}  –  \frac{f_{0}}{w²_{n}  –  w²}\right) \cos w_{n}t  +  \frac{f_{0}}{w²_{n}  –  w²} \cos wt          (2.11)

x(t) = \frac{f_{0}}{w²_{n}  –  w²} (\cos wt  –  \cos w_{n}t)        (2.12)

Using the trigonometric identity

cos u – cos ν = 2 \sin \left(\frac{ν  –  u}{2}\right) \sin \left(\frac{ν  +  u}{2}\right)

equation (2.12) becomes

x(t) = \frac{2f_{0}}{w²_{n}  –  w²} \sin \left(\frac{w_{n}  –  w}{2} t\right) \sin \left(\frac{w_{n}  +  w}{2} t\right)         (2.13)

The maximum value of the total response is evident from (2.13) so that

\frac{2f_{0}}{w²_{n}  –  w²} = 0.1 m

Solving this for m from w²_{n} = k/m and f_{0} = F_{0}/m yields

m = \frac{(0.1  m)(2000  N/m)  –  2(20  N)}{(0.1m)(10  ×  2 \pi  rad/s)²} = \frac{4}{\pi²} = 0.405 kg