The equation for energy (E) of a simple harmonic oscillator, `E = (1)/(2) mv^(2)+(1)/(2)m 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 factors involving mass (m), angular frequency (ω), and Planck's constant (h). Let's break down the solution step by step.

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

### Step 2: Identify Dimensions
We need to identify the dimensions of each term in the energy equation:
- The dimension of mass (m) is \([M]\).
- The dimension of velocity (v) is \([L][T^{-1}]\).
- The dimension of angular frequency (ω) is \([T^{-1}]\).
- The dimension of displacement (x) is \([L]\).

### Step 3: Analyze the Energy Terms
1. For the kinetic energy term \(\frac{1}{2} mv^2\):
\[ \text{Dimension} = [M][L^2][T^{-2}] = [M][L^2][T^{-2}] \]

2. For the potential energy term \(\frac{1}{2} m \omega^2 x^2\):
\[ \text{Dimension} = [M][T^{-2}][L^2] = [M][L^2][T^{-2}] \]

Both terms have the same dimension: \([M][L^2][T^{-2}]\).

### Step 4: Making the Equation Dimensionless
To make the energy dimensionless, we can multiply it by a suitable factor that has the inverse dimensions of energy.

### Step 5: Relate Planck's Constant
Planck's constant \( h \) has dimensions of:
\[ [h] = [M][L^2][T^{-1}] \]
We can relate \( h \) to the other quantities.

### Step 6: Find Suitable Factors
We need to find a combination of \( m \), \( \omega \), and \( h \) that gives us a dimensionless quantity.

1. The momentum \( p \) is given by:
\[ p = mv \]
where \( v = \omega r \) (velocity in terms of angular frequency and radius).

2. Thus, \( p = m(\omega r) \).

3. The dimensions of momentum \( p \) are:
\[ [p] = [M][L][T^{-1}] \]

4. We can express \( r \) in terms of \( h \) and \( m \) and \( \omega \):
\[ r^2 \sim \frac{h}{m \omega} \]
Taking the square root gives:
\[ r \sim \sqrt{\frac{h}{m \omega}} \]

### Conclusion
The unit of momentum is \([M][L][T^{-1}]\) and the unit of length is \(\sqrt{\frac{h}{m \omega}}\).