A) 0.6 s
B) 0.4 s
C) 0.3 s
D) 0.2 s
Correct Answer: B
Solution :
Let the displacement of the particle be given by \[x=A\sin \left( \frac{2\pi t}{T} \right)\] where A = 4 cm & T = 1.2 s. If \[{{t}_{1}}\]is the time taken by the particle to move from \[x=o\]to \[x=2\,cm,\]then\[2=4\sin \left( \frac{2\pi {{t}_{1}}}{T} \right)\]which gives \[{{t}_{1}}=T/12.\] If \[{{t}_{2}}\]is the time taken to move from \[x=0\] to \[x=4\,cm,\]then \[4=4\sin \left( \frac{2\pi {{t}_{2}}}{T} \right)\] which gives \[{{t}_{2}}=T/4.\] Therefore, time taken to move from \[x=2\,cm\]to \[x=4\,cm\]is \[{{t}_{2}}-{{t}_{1}}=\frac{T}{4}-\frac{T}{12}=\frac{T}{6}=\frac{1.2s}{6}=0.2\,\sec .\] \[\therefore \] time taken by the particle to move from \[x=2\,cm\]to\[x=4\,cm\] & back = 0.4 sec. Hence, the correction option is (b).
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