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 solve the problem of making the energy equation of a simple harmonic oscillator dimensionless, we need to determine the units of momentum and length based on the constants involved: mass (m), angular frequency (ω), and Planck's constant (h).

### Step-by-Step Solution:

1. **Identify the Energy Equation**:
The energy (E) of a simple harmonic oscillator is given by:
\[ E = \frac{1}{2} mv^2 + \frac{1}{2} kx^2 \]
where \( k = m\omega^2 \) for a harmonic oscillator.

2. **Express the Energy in Terms of Dimensions**:
The dimensions of each term in the energy equation can be expressed as follows:
- The dimension of mass \( m \) is \( [M] \).
- The dimension of velocity \( v \) is \( [L][T^{-1}] \).
- Therefore, the dimension of \( mv^2 \) is:
\[ [M][L^2][T^{-2}] \]
- The dimension of \( kx^2 \) (where \( k = m\omega^2 \)) is also \( [M][L^2][T^{-2}] \).

3. **Planck's Constant**:
The dimension of Planck's constant \( h \) is given by:
\[ [h] = [E][T] = [M][L^2][T^{-2}][T] = [M][L^2][T^{-1}] \]

4. **Determine the Dimensions of \( m\omega h \)**:
- The angular frequency \( \omega \) has dimensions of \( [T^{-1}] \).
- Therefore, the dimensions of \( m\omega h \) are:
\[ [M][T^{-1}][M][L^2][T^{-1}] = [M^2][L^2][T^{-2}] \]

5. **Momentum**:
The momentum \( p \) is defined as:
\[ p = mv \]
Thus, the dimensions of momentum are:
\[ [M][L][T^{-1}] \]

6. **Length**:
To find a suitable expression for length in terms of \( m \), \( \omega \), and \( h \), we can consider:
\[ L = \frac{h}{m\omega} \]
The dimensions of this expression can be derived as follows:
- The dimension of \( h \) is \( [M][L^2][T^{-1}] \).
- The dimension of \( m\omega \) is:
\[ [M][T^{-1}] \]
- Therefore, the dimensions of \( L \) become:
\[ \frac{[M][L^2][T^{-1}]}{[M][T^{-1}]} = [L^2] \]
- Taking the square root gives us the dimension of length \( [L] \).

### Final Expressions:
- The unit of momentum is \( [M][L][T^{-1}] \).
- The unit of length is \( \frac{h}{m\omega} \).