question_answer

An oscillator consists of a block attached to a spring (k = 400 N/m). At some time t, the position (measured from the systems equilibrium location), velocity and acceleration of the block are \[x=0.100\,m,\,v=-15.0\,m/s\], and\[a=-90\,m/{{s}^{2}}\]. The amplitude of the motion and the mass of the block are

A)  0.2 m, 0.84 kg

B)  0.3 m, 0.76 kg

C)  0.4 m, 0.54 kg

D)  0.5 m, 0.44 kg

Correct Answer: D

Solution :

                 Given that, spring constant K = 400 N/m                 Position y = 0.100 m                 Velocity V = - 15.0 m/s                 and acceleration a = 90 \[m/{{s}^{2}}\] We know that                                 \[v=\omega \sqrt{{{A}^{2}}-{{y}^{2}}}\]                                .... (i) and        \[a=-{{\omega }^{2}}y\]                                               ?. (ii) From Eq. (ii),                 \[90={{\omega }^{2}}\times 0.1\] \[\Rightarrow \]               \[\omega =30\] How       \[{{\omega }^{2}}=\frac{k}{m}\]                 \[9\omega =\frac{400}{m}\]                 \[m=\frac{4}{9}=0.44\,kg\] From Eq (i),                 \[15=\omega \sqrt{{{A}^{2}}-{{y}^{2}}}\]                 \[225=900({{A}^{2}}-{{y}^{2}})\]                 \[225=900{{A}^{2}}-900{{(0-1)}^{2}}\]                 \[{{A}^{2}}=\frac{234}{900}\]                 \[A=\frac{15}{30}=\frac{1}{2}=0.5\,m\]