Amplitude of oscillation of a particle that executes S.H.M. is 2 cm. Its displacement from its mean position in a time equal to 1/6th of its time period is
√3 cm
1/√3 cm
1/√2 cm
√2 cm
A
√2 cm
B
1/√3 cm
C
√3 cm
D
1/√2 cm
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Solution
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A=2cm x=Asinwt =Asin(2πT×T6)=A×√32 x=√3cm
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