A body executing `SHM` has a maximum velocity of `1ms^(-1)` and a maximum acceleration of `4ms^(-1)`. Its time period of oscillation is

Options

A. `3.14s`
B. `1.57s`
C. `6.28s`
D. `0.25s`

Explanation

To find the time period of a body executing Simple Harmonic Motion (SHM) given its maximum velocity and maximum acceleration, we can follow these steps:

### Step 1: Identify the given values
- Maximum velocity, \( V_{\text{max}} = 1 \, \text{m/s} \)
- Maximum acceleration, \( A_{\text{max}} = 4 \, \text{m/s}^2 \)

### Step 2: Use the relationship between maximum acceleration and maximum velocity
In SHM, the maximum acceleration is related to the maximum velocity and angular frequency (\( \omega \)) by the formula:
\[ A_{\text{max}} = V_{\text{max}} \cdot \omega \]
From this, we can express \( \omega \) as:
\[ \omega = \frac{A_{\text{max}}}{V_{\text{max}}} \]

### Step 3: Substitute the known values to find \( \omega \)
Substituting the given values into the equation:
\[ \omega = \frac{4 \, \text{m/s}^2}{1 \, \text{m/s}} = 4 \, \text{s}^{-1} \]

### Step 4: Relate angular frequency to the time period
The angular frequency \( \omega \) is related to the time period \( T \) by the formula:
\[ \omega = \frac{2\pi}{T} \]
We can rearrange this to find \( T \):
\[ T = \frac{2\pi}{\omega} \]

### Step 5: Substitute \( \omega \) to find the time period \( T \)
Now, substituting \( \omega = 4 \, \text{s}^{-1} \) into the equation for \( T \):
\[ T = \frac{2\pi}{4} = \frac{\pi}{2} \, \text{s} \]

### Step 6: Calculate the numerical value of \( T \)
Calculating the numerical value:
\[ T \approx 1.57 \, \text{s} \]

### Final Answer
The time period of oscillation is approximately \( 1.57 \, \text{s} \).

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