A spring is hanging down from the ceiling, and an object of mass m is attached to the free end. The object is pulled down, thereby stretching the spring, and then released. The object oscillates up and down, and the time T required for one complete up-and-down oscillation is given by the equation `T=2pisqrt(m//k)`, where k is known as the spring constant. What must be the dimension of k for this equation to be dimensionally correct?

Options

A. `([M])/([T])`
B. `([T])/([M])`
C. `([M])/([T]^(2))`
D. `([T])/([M]^(2))`

Answer

To determine the dimension of the spring constant \( k \) in the equation \( T = 2\pi \sqrt{\frac{m}{k}} \), we will follow these steps:

### Step 1: Identify the dimensions involved
- The time period \( T \) has the dimension of time, which is denoted as \( [T] \).
- The mass \( m \) has the dimension of mass, denoted as \( [M] \).
- The spring constant \( k \) is what we need to find the dimensions of.

### Step 2: Rearrange the equation to isolate \( k \)
The equation given is:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]
To isolate \( k \), we can first square both sides:
\[ T^2 = (2\pi)^2 \frac{m}{k} \]
Since \( (2\pi)^2 \) is a constant, we can ignore it for dimensional analysis:
\[ T^2 = \frac{m}{k} \]

### Step 3: Rearranging to solve for \( k \)
Now, rearranging the equation gives:
\[ k = \frac{m}{T^2} \]

### Step 4: Determine the dimensions of \( k \)
Now we can express the dimensions of \( k \):
- The dimension of mass \( m \) is \( [M] \).
- The dimension of time squared \( T^2 \) is \( [T^2] \).

Thus, the dimension of \( k \) can be expressed as:
\[ [k] = \frac{[M]}{[T^2]} = [M][T]^{-2} \]

### Step 5: Conclusion
Therefore, the dimension of the spring constant \( k \) is:
\[ k = [M][T]^{-2} \]

### Final Answer
The dimension of \( k \) is \( M T^{-2} \).

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