A particle of mass 2gm is initially displaced through 2cm and then released. The frictional force constant due to air on it is12×10−3N/m . The restoring force constant is50×10−3N/m . If it is in oscillatory motion, its time period is

Fs=FResultant; kl=(mg+f); l=mg+fk l=2×10−3×9.8+12×10−350×10−3 =(19.6+12)50×10−3=(19.6+12)10−350+10−3=0.52 m T=2πlg=2π0.529.8 T=2π(14)=π2sec

A particle of mass 2gm is initially displaced through 2cm and then released. The frictional force constant due to air on it is12×10−3N/m . The restoring force constant is50×10−3N/m . If it is in oscillatory motion, its time period is

Fs=FResultant; kl=(mg+f); l=mg+fk l=2×10−3×9.8+12×10−350×10−3 =(19.6+12)50×10−3=(19.6+12)10−350+10−3=0.52 m T=2πlg=2π0.529.8 T=2π(14)=π2sec

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A particle of mass 2gm is initially displaced through 2cm and then released. The frictional force constant due to air on it is $12 \times 10^{- 3} N/m$ . The restoring force constant is $50 \times 10^{- 3} N/m$ . If it is in oscillatory motion, its time period is

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π \sec

4 π \sec

π /2 \sec

2 π \sec

answer is B.

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F_{s} = F_{Resultant}; k l = (mg + f); l = \frac{mg + f}{k}

$l = \frac{2 \times 10^{- 3} \times 9.8 + 12 \times 10^{- 3}}{50 \times 10^{- 3}}$

$= \frac{(19.6 + 12)}{50 \times 10^{- 3}} = \frac{(19.6 + 12) 10^{- 3}}{50 + 10^{- 3}} = 0.52 m$

$T = 2 π \sqrt{\frac{l}{g}} = 2 π \sqrt{\frac{0.52}{9.8}}$

$T = 2 π (\frac{1}{4}) = \frac{π}{2} \sec$

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A particle of mass 2gm is initially displaced through 2cm and then released. The frictional force constant due to air on it is12×10−3N/m . The restoring force constant is50×10−3N/m . If it is in oscillatory motion, its time period is

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A particle of mass 2gm is initially displaced through 2cm and then released. The frictional force constant due to air on it is $12 \times 10^{- 3} N/m$ . The restoring force constant is $50 \times 10^{- 3} N/m$ . If it is in oscillatory motion, its time period is