The angle made by the string of a simple pendulum with the vertical depends on time as `theta=pi/90sin[(pis^-1)t]`. Find the length of the pendulum if `g=pi^2ms^-2`

Explanation

To solve the problem, we need to find the length of the pendulum given the angle as a function of time and the acceleration due to gravity. Let's break it down step by step.

### Step 1: Identify the given information
We are given the angle \( \theta \) as:
\[ \theta = \frac{\pi}{90} \sin\left(\pi t^{-1}\right) \]
and the value of \( g \) as:
\[ g = \pi^2 \, \text{m/s}^2 \]

### Step 2: Determine the angular frequency
From the equation of motion, we can identify the angular frequency \( \omega \) from the sine function. The general form of simple harmonic motion is:
\[ \theta(t) = \theta_0 \sin(\omega t) \]
Comparing this with our equation, we find:
\[ \omega = \pi \]

### Step 3: Calculate the time period
The time period \( T \) of a simple pendulum is given by:
\[ T = \frac{2\pi}{\omega} \]
Substituting the value of \( \omega \):
\[ T = \frac{2\pi}{\pi} = 2 \, \text{seconds} \]

### Step 4: Relate the time period to the length of the pendulum
The time period \( T \) of a simple pendulum is also related to its length \( l \) and the acceleration due to gravity \( g \) by the formula:
\[ T = 2\pi \sqrt{\frac{l}{g}} \]
Substituting the known value of \( T \):
\[ 2 = 2\pi \sqrt{\frac{l}{g}} \]

### Step 5: Simplify the equation
Dividing both sides by \( 2 \):
\[ 1 = \pi \sqrt{\frac{l}{g}} \]
Now, squaring both sides:
\[ 1 = \pi^2 \frac{l}{g} \]

### Step 6: Solve for the length \( l \)
Rearranging the equation gives:
\[ l = \frac{g}{\pi^2} \]
Substituting the value of \( g \):
\[ l = \frac{\pi^2}{\pi^2} = 1 \, \text{meter} \]

### Final Answer
The length of the pendulum is:
\[ l = 1 \, \text{meter} \]