A simple pendulum attached to the celling of a stationary lift has time period T. The lift starts moving and the position vector of the lift is given as `vecr(t)=15t^2hati+15t^2hatj`, where r(t) is in meters and t is in seconds. If `g=10 m//s^2`, the new time period of oscillation is

Options

A. `T/sqrt(18)`
B. `T/3`
C. `T/sqrt5`
D. `T/sqrt6`

Answer

To solve the problem of finding the new time period of a simple pendulum in a moving lift, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Initial Condition**:
- The time period of a simple pendulum at rest is given by the formula:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
- Here, \(L\) is the length of the pendulum and \(g = 10 \, \text{m/s}^2\).

2. **Finding the Position Vector**:
- The position vector of the lift is given as:
\[ \vec{r}(t) = 15t^2 \hat{i} + 15t^2 \hat{j} \]
- This indicates that the lift is moving in both the x and y directions with respect to time.

3. **Calculating the Velocity**:
- The velocity \(\vec{v}(t)\) is the derivative of the position vector with respect to time:
\[ \vec{v}(t) = \frac{d\vec{r}}{dt} = \frac{d}{dt}(15t^2 \hat{i} + 15t^2 \hat{j}) = 30t \hat{i} + 30t \hat{j} \]

4. **Calculating the Acceleration**:
- The acceleration \(\vec{a}(t)\) is the derivative of the velocity with respect to time:
\[ \vec{a}(t) = \frac{d\vec{v}}{dt} = \frac{d}{dt}(30t \hat{i} + 30t \hat{j}) = 30 \hat{i} + 30 \hat{j} \]

5. **Determining the Pseudo Force**:
- In a non-inertial frame (the moving lift), a pseudo force acts on the pendulum. The pseudo force \(\vec{F}_{\text{pseudo}}\) is equal to:
\[ \vec{F}_{\text{pseudo}} = -m\vec{a} = -m(30 \hat{i} + 30 \hat{j}) \]
- This pseudo force acts downward, effectively increasing the gravitational force experienced by the pendulum.

6. **Calculating Effective Gravity**:
- The effective gravitational acceleration \(g_{\text{effective}}\) is the vector sum of the actual gravitational acceleration and the pseudo force's acceleration:
\[ g_{\text{effective}} = g + |\vec{a}| = 10 + \sqrt{30^2 + 30^2} \]
- Calculate the magnitude:
\[ |\vec{a}| = \sqrt{30^2 + 30^2} = \sqrt{900 + 900} = \sqrt{1800} = 30\sqrt{2} \]
- Therefore,
\[ g_{\text{effective}} = 10 + 30\sqrt{2} \]

7. **Finding the New Time Period**:
- The new time period \(T'\) of the pendulum in the moving lift can be expressed as:
\[ T' = 2\pi \sqrt{\frac{L}{g_{\text{effective}}}} \]
- To find the ratio of the new time period to the original time period:
\[ \frac{T'}{T} = \frac{2\pi \sqrt{\frac{L}{g_{\text{effective}}}}}{2\pi \sqrt{\frac{L}{g}}} = \sqrt{\frac{g}{g_{\text{effective}}}} \]
- Substituting \(g = 10\) and \(g_{\text{effective}} = 10 + 30\sqrt{2}\):
\[ T' = T \sqrt{\frac{10}{10 + 30\sqrt{2}}} \]

### Final Answer:
The new time period \(T'\) of the pendulum in the moving lift is:
\[ T' = T \sqrt{\frac{10}{10 + 30\sqrt{2}}} \]