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), \[15=\omega \sqrt{{{A}^{2}}-{{y}^{2}}}\] \[225=900({{A}^{2}}-{{y}^{2}})\] \[225=900{{A}^{2}}-900{{(0-1)}^{2}}\] \[{{A}^{2}}=\frac{234}{900}\] \[A=\frac{15}{30}=\frac{1}{2}=0.5\,m\]You need to login to perform this action.
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