A simple pendulum of length 5 m is suspended from the ceiling of a cart. Cart is sliding down on a frictionless surface having angle of inclination 60°. The time period of the pendulum is

Options

A. `2 pi s`
B. `pi s`
C. `4pi s`
D. `pi/2 s`

Answer

To find the time period of a simple pendulum suspended from a cart sliding down an inclined plane, we can follow these steps:

### Step 1: Identify the parameters
We have a pendulum of length \( L = 5 \, \text{m} \) and the cart is sliding down an inclined plane at an angle of \( \theta = 60^\circ \).

### Step 2: Understand the effective gravitational acceleration
The effective gravitational acceleration \( g_{\text{effective}} \) acting on the pendulum needs to be determined. Since the cart is accelerating down the incline, we need to consider the component of gravitational acceleration acting along the direction of the pendulum's mean position.

### Step 3: Calculate \( g_{\text{effective}} \)
The gravitational acceleration \( g \) is directed downward and has a value of approximately \( 10 \, \text{m/s}^2 \).

1. The angle between the vertical (downward direction) and the direction of the pendulum's mean position is \( 30^\circ \) (since \( 90^\circ - 60^\circ = 30^\circ \)).
2. The effective gravitational acceleration can be calculated using the cosine of the angle:
\[ g_{\text{effective}} = g \cos(60^\circ) \]
\[ g_{\text{effective}} = 10 \cdot \frac{1}{2} = 5 \, \text{m/s}^2 \]

### Step 4: Use the formula for the time period of a pendulum
The time period \( T \) of a simple pendulum is given by the formula:
\[ T = 2\pi \sqrt{\frac{L}{g_{\text{effective}}}} \]

### Step 5: Substitute the values into the formula
Substituting \( L = 5 \, \text{m} \) and \( g_{\text{effective}} = 5 \, \text{m/s}^2 \):
\[ T = 2\pi \sqrt{\frac{5}{5}} = 2\pi \sqrt{1} = 2\pi \]

### Step 6: Final answer
Thus, the time period of the pendulum is:
\[ T = 2\pi \, \text{seconds} \]

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