Question

The amplitude of a simple pendulum, oscillating in air with a small spherical bob, decreases from $10\, cm$ to $8\, cm$ 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 $10\, cm$ to $5 \,cm$ carbon dioxide will be close to ($\ell n\, 5 \,= \,1.601, \ell n\, 2\, =\, 0.693)$.

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

Answer 1

The Correct Option is C

To solve this problem, we'll use the concept of damping in a simple pendulum and apply Stokes' law, which is relevant here since the pendulum is swinging in fluids (air and carbon dioxide).

The amplitude of the oscillation decreases exponentially over time, given by:

\(A(t) = A_0 e^{-\beta t}\),

where \(A_0\) is the initial amplitude and \(\beta\) is the damping constant that depends on the medium's viscosity.

Given that the amplitude decreases from \(10 \, \text{cm}\) to \(8 \, \text{cm}\) in \(40 \, s\) in air, we can write:

\(8 = 10 e^{-\beta_{\text{air}} \times 40}\)

Rearranging for \(\beta_{\text{air}}\) gives:

\(\beta_{\text{air}} = \frac{\ln(10/8)}{40} = \frac{\ln(1.25)}{40}\)

The damping constant \(\beta\) is proportional to the viscosity. According to the problem, the ratio of the coefficient of viscosity of air to that of carbon dioxide is \(1.3\). Thus,

\(\beta_{\text{CO}_2} = \frac{\beta_{\text{air}}}{1.3}\)

Now, for the pendulum in carbon dioxide, the amplitude decreases from \(10 \, \text{cm}\) to \(5 \, \text{cm}\). We set up the equation:

\(5 = 10 e^{-\beta_{\text{CO}_2} \times t_{\text{CO}_2}}\)

Rearranging gives:

\(t_{\text{CO}_2} = \frac{\ln(10/5)}{\beta_{\text{CO}_2}} = \frac{\ln(2)}{\beta_{\text{CO}_2}}\)

Substitute the value of \(\beta_{\text{CO}_2}\):

\(t_{\text{CO}_2} = \frac{\ln(2) \times 1.3}{\ln(1.25)/40}\)

Calculating, we have:

\(\ln(1.25) \approx 0.2231 \, (\text{using Taylor expansion or calculator})\)

\(\ln(2) = 0.693\)

Therefore, the time \(t_{\text{CO}_2}\) is:

\(t_{\text{CO}_2} = \frac{0.693 \times 1.3 \times 40}{0.2231} \approx 160.62 \, \text{seconds}\)

Hence, the closest answer is 161 seconds.

Therefore, the correct answer is:

Option: 161 s