A) 10cm
B) 9.5cm
C) 4cm
D) 11.36cm
Correct Answer: B
Solution :
Key Idea: Length of path is twice of amplitude. The velocity (u) of a body in SHM changes with displacement (y) as follows \[u=\omega \sqrt{{{a}^{2}}-{{y}^{2}}}\] where a is amplitude and co is angular speed. Given, \[{{u}_{1}}=10\,cm/s,\,{{u}_{2}}=7\,cm/s\] \[{{y}_{1}}=3\,cm,\,{{y}_{2}}=4\,cm\] \[\therefore \] \[\frac{{{u}_{2}}}{{{u}_{1}}}=\frac{7}{10}=\sqrt{\frac{{{a}^{2}}-16}{{{a}^{2}}-9}}\] \[\Rightarrow \] \[a=4.77\,cm\] Length of path \[=2\,a=2\times 4.77\approx 9.5\]You need to login to perform this action.
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