Find the velocity when KE = PE of the body undergoing SHM. Amplitude = `x_(0)` and angular frequency is `omega` . How many times in a cycle KE = PE ?

Options

A. `(omegax_(0))/(sqrt(2))`,2
B. `omegax_(0)` ,2
C. `(omegax_(0))/(sqrt(2))`,4
D. `omegax_(0)`,4

Answer

To solve the problem, we need to find the velocity of a body undergoing Simple Harmonic Motion (SHM) when its kinetic energy (KE) equals its potential energy (PE). We also need to determine how many times this occurs in one complete cycle of motion.

### Step-by-Step Solution:

1. **Understanding SHM Parameters**:
- The amplitude of the motion is given as \( x_0 \).
- The angular frequency is given as \( \omega \).

2. **Kinetic Energy and Potential Energy in SHM**:
- The kinetic energy (KE) of a body in SHM is given by:
\[ KE = \frac{1}{2} m v^2 \]
- The potential energy (PE) in SHM is given by:
\[ PE = \frac{1}{2} k x^2 \]
- Here, \( k \) is the spring constant, which can be expressed in terms of angular frequency as \( k = m \omega^2 \).

3. **Setting KE Equal to PE**:
- We need to find the position \( x \) where \( KE = PE \):
\[ \frac{1}{2} m v^2 = \frac{1}{2} k x^2 \]
- Substituting \( k \):
\[ \frac{1}{2} m v^2 = \frac{1}{2} m \omega^2 x^2 \]
- Canceling \( \frac{1}{2} m \) from both sides (assuming \( m \neq 0 \)):
\[ v^2 = \omega^2 x^2 \]

4. **Finding Velocity**:
- The velocity \( v \) can be expressed in terms of displacement \( x \):
\[ v = \omega \sqrt{x_0^2 - x^2} \]
- Since we need to find \( v \) when \( KE = PE \), we substitute \( x^2 \) from the previous equation:
\[ v = \omega \sqrt{x_0^2 - \left(\frac{x_0}{\sqrt{2}}\right)^2} \]
- This simplifies to:
\[ v = \omega \sqrt{x_0^2 - \frac{x_0^2}{2}} = \omega \sqrt{\frac{x_0^2}{2}} = \frac{\omega x_0}{\sqrt{2}} \]

5. **Finding How Many Times KE = PE in One Cycle**:
- The condition \( KE = PE \) occurs when the displacement \( x = \pm \frac{x_0}{\sqrt{2}} \).
- In one complete cycle of SHM, the particle will pass through this position twice (once while moving in the positive direction and once while moving in the negative direction).
- Therefore, the total occurrences in one cycle are:
\[ \text{Total occurrences} = 4 \text{ (2 times in positive and 2 times in negative direction)} \]

### Final Answers:
- The velocity when \( KE = PE \) is:
\[ v = \frac{\omega x_0}{\sqrt{2}} \]
- The number of times in one cycle that \( KE = PE \) is:
\[ 4 \text{ times} \]