Question

The amplitude of a simple pendulum, oscillating in air with a small spherical bob, decreases from $10cm$ to $8cm$ in $40s$. 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 $(ln5=1.601,ln2=0.693)$.

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

Answer 1

The correct option is

B

$161s$

In air, $10cm\underrightarrow{40s}8cm$

In $C{O}_2$, $10cm\underrightarrow{?}5cm$

$\because \frac{n_a}{n_{{CO}_2}}=1.3$

We know that the amplitude of damped oscillation is given by,

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

$\Rightarrow 8=10e^{(-\frac{b_{\text{air}}\times 40}{2m})}\text{and}5=10e^{-\left(\frac{b_{c{o}_2}\times t_2}{2m}\right)}$

$\Rightarrow \frac{\mathrm{In}\frac{5}{4}}{\mathrm{In}2}=\frac{b_{\text{air}}\times 40}{b_{c{o}_2}\times t_2}$.

Therefore,

$t_2=\frac{b_{air}}{b_{co}}\times \frac{40\times \mathrm{In}2}{\mathrm{In}\frac{5}{4}}=161s$.