A) \[-4\]
B) 4
C) \[4\sqrt{2}\]
D) 8
Correct Answer: C
Solution :
We have the displacement of the particle in SHM given by \[x=4(cos\pi t+\sin \pi t)\] \[x=4\cos \pi t+4sin\,\pi t\] From the given equation \[{{a}_{1}}=4\] and \[{{a}_{2}}=4\] \[\therefore \] amplitude \[a=\sqrt{a_{1}^{2}+a_{2}^{2}}\] \[=\sqrt{{{4}^{2}}+{{4}^{2}}}\] \[a=4\sqrt{2}\]You need to login to perform this action.
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