A seconds pendulum is suspended from rof of a vehicle that is moving along a circular track of radius `(10)/(sqrt(3))m` with speed `10m//s`. Its period of oscillation will be `(g = 10m//s^(2))`

Options

A. `sqrt(2)s`
B. `2s`
C. `1s`
D. `0.5s`

Explanation

To solve the problem of finding the period of oscillation of a seconds pendulum suspended from the roof of a vehicle moving along a circular track, we can follow these steps:

### Step 1: Understand the Problem
We have a seconds pendulum (which has a period of 2 seconds) suspended from a vehicle that is moving in a circular path. The radius of the circular track is given as \( r = \frac{10}{\sqrt{3}} \) m, and the speed of the vehicle is \( v = 10 \) m/s. We need to find the period of oscillation of the pendulum when the effective acceleration due to gravity is considered.

### Step 2: Identify the Effective Gravity
When the vehicle is moving in a circular path, the pendulum experiences both gravitational force and centripetal acceleration. The effective acceleration \( g' \) acting on the pendulum can be expressed as:
\[ g' = g + a_c \]
where \( a_c \) is the centripetal acceleration given by:
\[ a_c = \frac{v^2}{r} \]
Substituting the values:
- \( g = 10 \, \text{m/s}^2 \)
- \( v = 10 \, \text{m/s} \)
- \( r = \frac{10}{\sqrt{3}} \)

### Step 3: Calculate Centripetal Acceleration
Now, we calculate \( a_c \):
\[ a_c = \frac{v^2}{r} = \frac{10^2}{\frac{10}{\sqrt{3}}} = \frac{100}{\frac{10}{\sqrt{3}}} = 10\sqrt{3} \, \text{m/s}^2 \]

### Step 4: Calculate Effective Gravity
Now, substituting \( a_c \) back into the equation for effective gravity:
\[ g' = g + a_c = 10 + 10\sqrt{3} \]

### Step 5: Determine the Length of the Pendulum
For a seconds pendulum, the length \( L \) can be expressed as:
\[ L = \frac{g}{\pi^2} \]

### Step 6: Calculate the Period of the Pendulum
The period \( T \) of the pendulum is given by:
\[ T = 2\pi \sqrt{\frac{L}{g'}} \]
Substituting \( L \) and \( g' \):
\[ T = 2\pi \sqrt{\frac{\frac{g}{\pi^2}}{g + 10\sqrt{3}}} \]
This simplifies to:
\[ T = 2\pi \sqrt{\frac{g}{\pi^2(g + 10\sqrt{3})}} \]

### Step 7: Substitute the Values
Substituting \( g = 10 \):
\[ T = 2\pi \sqrt{\frac{10}{\pi^2(10 + 10\sqrt{3})}} = 2\pi \sqrt{\frac{10}{10\pi^2(1 + \sqrt{3})}} = 2\sqrt{\frac{1}{\pi^2(1 + \sqrt{3})}} \]

### Step 8: Final Calculation
Now, we can calculate the numerical value of \( T \):
\[ T = \frac{2}{\pi} \sqrt{\frac{1}{1 + \sqrt{3}}} \]
This gives us the final period of oscillation of the pendulum.

### Final Answer
The period of oscillation of the seconds pendulum is approximately \( \sqrt{2} \) seconds.

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