The equation for energy (E) of a simple harmonic oscillator, `E = (1)/(2) mv^(2)+(1)/(2)+ omega^(2)x^(2)`?, is to be made "dimensionless" by multiplying by a suitable factor, which may involve the constants, m(mass), `omega`(angular frequency) and h(Planck's constant). What will be the unit of momentum and length?

Answer

To make the energy equation of a simple harmonic oscillator dimensionless, we need to find suitable combinations of the constants mass (m), angular frequency (ω), and Planck's constant (h).

### Step-by-Step Solution:

1. **Understanding the Energy Equation**:
The energy (E) of a simple harmonic oscillator is given by:
\[ E = \frac{1}{2} mv^2 + \frac{1}{2} \omega^2 x^2 \]
Here, \(v\) is the velocity and \(x\) is the displacement.

2. **Identifying Dimensions**:
- The dimension of mass \(m\) is \([M]\).
- The dimension of angular frequency \(\omega\) is \([T^{-1}]\) (since \(\omega = \frac{2\pi}{T}\)).
- The dimension of Planck's constant \(h\) is \([M L^2 T^{-1}]\) (since \(h\) has dimensions of energy multiplied by time).

3. **Finding the Dimension of Momentum**:
The momentum \(p\) is defined as:
\[ p = mv \]
The dimension of velocity \(v\) is \([L T^{-1}]\). Therefore, the dimension of momentum is:
\[ [p] = [M][L T^{-1}] = [M L T^{-1}] \]

4. **Finding the Dimension of Length**:
To express length in terms of \(m\), \(\omega\), and \(h\), we can use the following relationship:
\[ L = \sqrt{\frac{h}{m \omega}} \]
This implies that the dimension of length can be derived from the combination of these constants.

5. **Final Expressions**:
- The unit of momentum is \( [M L T^{-1}] \).
- The unit of length can be expressed as:
\[ [L] = \sqrt{\frac{h}{m \omega}} \]

### Summary:
- **Unit of Momentum**: \([M L T^{-1}]\)
- **Unit of Length**: \(\sqrt{\frac{h}{m \omega}}\)