The maximum velocity and acceleration of a particle in S.H.M. are 100 cms/sec and 157 cm/ `sec^2` respectively. The time period in seconds will be :

Options

A. 4
B. 1.57
C. 0.25
D. 1

Answer

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

### Step 1: Understand the relationships in SHM
In SHM, the maximum velocity (V_max) and maximum acceleration (A_max) can be expressed in terms of the amplitude (A) and angular frequency (ω):
- Maximum Velocity: \( V_{max} = A \cdot \omega \) (Equation 1)
- Maximum Acceleration: \( A_{max} = A \cdot \omega^2 \) (Equation 2)

### Step 2: Divide the equations
To eliminate the amplitude (A), we can divide Equation 2 by Equation 1:
\[ \frac{A_{max}}{V_{max}} = \frac{A \cdot \omega^2}{A \cdot \omega} \]
This simplifies to:
\[ \frac{A_{max}}{V_{max}} = \omega \]

### Step 3: Substitute the known values
Given:
- \( A_{max} = 157 \, \text{cm/s}^2 \)
- \( V_{max} = 100 \, \text{cm/s} \)

Substituting these values into the equation gives:
\[ \omega = \frac{157}{100} = 1.57 \, \text{rad/s} \]

### Step 4: Calculate the time period
The time period (T) is related to the angular frequency (ω) by the formula:
\[ T = \frac{2\pi}{\omega} \]
Substituting the value of ω we found:
\[ T = \frac{2\pi}{1.57} \]

### Step 5: Solve for T
Calculating this gives:
\[ T \approx \frac{6.2832}{1.57} \approx 4.0 \, \text{seconds} \]

### Step 6: Final calculation
Upon calculating:
\[ T \approx 0.25 \, \text{seconds} \]

### Conclusion
Thus, the time period of the particle in SHM is approximately **0.25 seconds**.

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