| The oscillations represented by curve 1 in the graph are expressed by equation \[x=A\sin \omega \,t.\] The equation for the oscillations represented by curve 2 is expressed as: |
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A) \[x=2A\sin \,(\omega \,t-\pi /2)\]
B) \[x=2A\sin \,(\omega \,t+\pi /2)\]
C) \[x=-\,2A\sin \,(\omega \,t-\pi /2)\]
D) \[x=A\sin \,(\omega \,t-\pi /2)\]
Correct Answer: A
Solution :
Oscillations represented by curve 2 lags in phase by \[\pi /2\] and the periods are same. Amplitude of curve 2 is double that of 1. Hence [A].
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