If the period of S.H.M. is 12 seconds and amplitude 10 cm. What is the displacement 14 Seconds after passage of the particle through its extreme positive displacement?
A
5 cm
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B
4 cm
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C
2 cm
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D
10 cm
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Solution
The correct option is A 5 cm The motion starts from extreme position, y=Acosωt=Acos(2πT)t =10cos[(2π12)×14]=10cos7π3 =10cosπ3=10×12=5cm.
Position from mean position is 5 cm hence the total displacement from the positive extreme position is 5 cm