Question

The amplitude of a simple pendulum, oscillating in air with a small spherical bob, decreases from $10cm$ to $8cm$ in $40$ seconds. Assuming that Stokes law is valid, and ratio of the coefficient of viscosity of air to that of carbon dioxide is $1.3$, the time In which amplitude of this pendulum will reduce from $10cm$ to $5cm$ in carbon dioxide will be close to (ln $5=1.601,\ln 22=0.693$)

• A. $231s$
• B. $208s$
• C. $161s$
• D. $142s$

Answer 1

Answer: (C)

$161s$

$\text{Solution}:$

We know that the amplitude of damped oscillation is

$A=A_0{e}^{\frac{-bt}{2m}}$

$⟹8=10e^{(-\frac{b_{air}\times 40}{2m})}$

and

$5=10e^{-\left(\frac{b_{c{o}_2}\times t_2}{2m}\right)}$

$⟹\frac{In\frac{5}{4}}{In2}=\frac{b_{air}\times 40}{b_{c{o}_2}\times t_2}$

Therefore ,

$t_2=\frac{b_{air}}{b_{co}}\times \frac{40\times In2}{In\frac{5}{4}}=161s$

$\text{Hence C is the correct option}$