If the speed v of a particle of mass m as function of time t is given by `v=omegaAsin[(sqrt(k)/(m))t]`, where A has dimension of length.

Options

A. the argument of trigonometric function must be a dimensionless quntity
B. Dimensional formula of `omega` is `LT^-1`
C. Dimensional formula of k is `MLT^-2`
D. Dimensional formula of `sqrt((k)/(m))` is T

Explanation

To solve the problem, we need to analyze the given equation for the speed \( v \) of a particle as a function of time \( t \):

\[ v = \omega A \sin\left(\sqrt{\frac{k}{m}} t\right) \]

where:
- \( \omega \) is a constant,
- \( A \) has the dimension of length,
- \( k \) is a constant related to the system,
- \( m \) is the mass of the particle.

### Step 1: Determine the dimensions of the argument of the sine function

The argument of the sine function, \( \sqrt{\frac{k}{m}} t \), must be dimensionless because the sine function only accepts dimensionless quantities.

To ensure this, we need to analyze the dimensions of \( k \) and \( m \):

- The dimension of mass \( m \) is \( [M] \).
- The dimension of \( k \) can be determined from its physical context (often related to spring constant or similar), which typically has dimensions of \( [M][L][T^{-2}] \).

Thus, we can write:

\[ \text{Dimension of } k = [M][L][T^{-2}] \]

### Step 2: Calculate the dimensions of \( \sqrt{\frac{k}{m}} \)

Now, we calculate the dimensions of \( \sqrt{\frac{k}{m}} \):

\[ \frac{k}{m} = \frac{[M][L][T^{-2}]}{[M]} = [L][T^{-2}] \]

Taking the square root gives:

\[ \sqrt{\frac{k}{m}} = \sqrt{[L][T^{-2}]} = [L^{1/2}][T^{-1}] \]

### Step 3: Combine with time \( t \)

Now, we multiply this by time \( t \):

\[ \sqrt{\frac{k}{m}} t = [L^{1/2}][T^{-1}] \cdot [T] = [L^{1/2}] \]

This shows that the argument \( \sqrt{\frac{k}{m}} t \) is not dimensionless. Therefore, we need to check the dimensions more carefully.

### Step 4: Ensure the argument is dimensionless

For \( \sqrt{\frac{k}{m}} t \) to be dimensionless, we need:

\[ \sqrt{\frac{k}{m}} \cdot t = \text{dimensionless} \]

This implies:

\[ \sqrt{\frac{k}{m}} \text{ must have dimensions of } [T^{-1}] \]

Thus, we conclude that:

\[ \frac{k}{m} \text{ must have dimensions of } [T^{-2}] \]

### Step 5: Final conclusions

1. The argument of the sine function must be dimensionless.
2. The dimensions of \( \omega \) can be derived from the equation \( v = \omega A \) where \( v \) has dimensions \( [L][T^{-1}] \) and \( A \) has dimensions \( [L] \). Thus, \( \omega \) must have dimensions \( [T^{-1}] \).
3. The dimensions of \( k \) can be found to be \( [M][L][T^{-2}] \), but we need to ensure that \( \sqrt{\frac{k}{m}} \) gives a dimensionless argument when multiplied by time.

### Summary of Options

- **Option A**: Correct - The argument of a trigonometric function must be dimensionless.
- **Option B**: Incorrect - The dimension of \( \omega \) is not \( [L][T^{-1}] \).
- **Option C**: Incorrect - The dimension of \( k \) is not \( [M][L][T^{-2}] \) in the context needed.
- **Option D**: Incorrect - The dimension of \( \sqrt{\frac{k}{m}} \) is not \( [T^{-1}] \) as required.