A mass 𝑚, suspended by a spring of stiffness 𝑘, is set into simple harmonic motion. The period of oscillation is given by:
t=2π√(m/k)

Using dimensional analysis, demonstrate that the period of oscillation does, indeed, have units of time.

To show that the period of oscillation \( t = 2\pi \sqrt{\frac{m}{k}} \) has units of time using dimensional analysis, we need to analyze the dimensions of the variables involved and verify that the dimensions on both sides of the equation are consistent with time.

### Step 1: Identify the dimensions of each variable.

- \( t \): This is the period of oscillation, which represents time. Hence, the dimension of \( t \) is:
  \[
  [t] = T
  \]
  where \( T \) represents the dimension of time.

- \( m \): This is the mass of the object, so the dimension of \( m \) is:
  \[
  [m] = M
  \]
  where \( M \) represents the dimension of mass.

- \( k \): This is the stiffness (spring constant) of the spring. From Hooke’s law, we know that \( F = kx \), where \( F \) is force and \( x \) is displacement. Therefore, the dimension of \( k \) can be derived as follows:
  \[
  k = \frac{F}{x}
  \]
  The dimension of force \( F \) is \( M L T^{-2} \), and the dimension of displacement \( x \) is \( L \). Therefore, the dimension of \( k \) is:
  \[
  [k] = \frac{M L T^{-2}}{L} = M T^{-2}
  \]

### Step 2: Analyze the formula \( t = 2\pi \sqrt{\frac{m}{k}} \).

We can ignore the constant \( 2\pi \), as it is dimensionless. Now, let's focus on the expression \( \sqrt{\frac{m}{k}} \).

- The dimension of \( \frac{m}{k} \) is:
  \[
  \left[ \frac{m}{k} \right] = \frac{M}{M T^{-2}} = T^2
  \]

- Therefore, the dimension of \( \sqrt{\frac{m}{k}} \) is:
  \[
  \left[ \sqrt{\frac{m}{k}} \right] = \sqrt{T^2} = T
  \]

### Step 3: Verify the dimensions.

From the analysis, we found that the dimension of \( \sqrt{\frac{m}{k}} \) is \( T \). Hence, the dimension of \( t \) (the period of oscillation) is also \( T \), which confirms that \( t \) has units of time.

### Conclusion:
The dimensional analysis verifies that the formula \( t = 2\pi \sqrt{\frac{m}{k}} \) is dimensionally consistent, and the period of oscillation has the correct units of time.

---

Would you like more details or have any other questions?

Here are five related questions you might find interesting:
1. How does the period change if the mass \( m \) is doubled?
2. How does the period depend on the spring stiffness \( k \)?
3. What would happen to the period if the spring is replaced by one with half the stiffness?
4. How does the amplitude of oscillation affect the period?
5. Can this formula be applied to a pendulum? Why or why not?

**Tip:** Dimensional analysis is a powerful tool to verify formulas and can help in deriving relationships even when exact values or constants are unknown.

A mass 𝑚, suspended by a spring of stiffness 𝑘, is set into simple harmonic motion. The period of oscillation is given by: t = 2π√(m/k). Using dimensional analysis, demonstrate that the period of oscillation does, indeed, have units of time.

Explore how dimensional analysis confirms that the period of oscillation in simple harmonic motion, t = 2π√(m/k), has units of time. This solution covers key concepts in oscillation, Hooke's law, and dimensional consistency. Suitable for students in Grades 11-12.

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