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A particle moves such that its acclerati...

A particle moves such that its accleration a is given by `a = -bx`, where x is the displacement from equilibrium position and b is a constant. The period of oscillation is

A

`sqrt(2pi) //b`

B

`2pi//sqrt(b)`

C

`2pi//b`

D

`2sqrt((pi//b))`

Text Solution

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The correct Answer is:
To solve the problem, we need to determine the period of oscillation for a particle whose acceleration is given by the equation \( a = -bx \), where \( x \) is the displacement from the equilibrium position and \( b \) is a constant. ### Step-by-Step Solution: 1. **Understanding the Given Equation**: The acceleration \( a \) is given by \( a = -bx \). This indicates that the acceleration is directly proportional to the displacement \( x \) but acts in the opposite direction (hence the negative sign). This is characteristic of simple harmonic motion (SHM). 2. **Identifying the Nature of Motion**: Since the acceleration is proportional to the displacement and directed towards the equilibrium position, we can conclude that the particle is undergoing simple harmonic motion (SHM). 3. **Using the Formula for Period in SHM**: The formula for the period \( T \) of a simple harmonic oscillator is given by: \[ T = 2\pi \sqrt{\frac{m}{k}} \] where \( m \) is the mass of the particle and \( k \) is the spring constant. In our case, we can relate the given acceleration to the standard form of SHM. 4. **Relating Acceleration to SHM**: In SHM, we have: \[ a = -\omega^2 x \] where \( \omega \) is the angular frequency. Comparing this with our equation \( a = -bx \), we can identify: \[ \omega^2 = b \] 5. **Finding the Period**: The angular frequency \( \omega \) is related to the period \( T \) by the formula: \[ T = \frac{2\pi}{\omega} \] Substituting \( \omega = \sqrt{b} \) into the period formula gives us: \[ T = \frac{2\pi}{\sqrt{b}} \] 6. **Final Expression for the Period**: Therefore, the period of oscillation \( T \) is: \[ T = 2\pi \sqrt{\frac{1}{b}} \] ### Conclusion: The period of oscillation for the particle is given by: \[ T = 2\pi \sqrt{\frac{1}{b}} \]

To solve the problem, we need to determine the period of oscillation for a particle whose acceleration is given by the equation \( a = -bx \), where \( x \) is the displacement from the equilibrium position and \( b \) is a constant. ### Step-by-Step Solution: 1. **Understanding the Given Equation**: The acceleration \( a \) is given by \( a = -bx \). This indicates that the acceleration is directly proportional to the displacement \( x \) but acts in the opposite direction (hence the negative sign). This is characteristic of simple harmonic motion (SHM). 2. **Identifying the Nature of Motion**: ...
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