To find the value of \( k \), we can rearrange the formula given:

\[ T = 2\pi \sqrt{\frac{m}{k}} \]

Given:
- \( T = 3.70 \)
- \( m = 67 \)

First, isolate \( \sqrt{\frac{m}{k}} \):

\[ \frac{T}{2\pi} = \sqrt{\frac{m}{k}} \]

Square both sides to remove the square root:

\[ \left(\frac{T}{2\pi}\right)^2 = \frac{m}{k} \]

Rearrange to solve for \( k \):

\[ k = \frac{m}{\left(\frac{T}{2\pi}\right)^2} \]

Substitute the given values:

\[ k = \frac{67}{\left(\frac{3.70}{2\pi}\right)^2} \]

Calculate the denominator:

\[ \frac{3.70}{2\pi} = \frac{3.70}{6.2832} \approx 0.5887 \]

Square this value:

\[ (0.5887)^2 \approx 0.3466 \]

Finally, compute \( k \):

\[ k = \frac{67}{0.3466} \approx 193.26 \]

So, the value of \( k \) is approximately \( 193.26 \, \text{N/m} \).

Would you like further details or have any questions about this calculation?

Here are 5 related questions to expand your understanding:

1. How does the period \( T \) change if the mass \( m \) is doubled?
2. What is the physical interpretation of the constant \( k \) in this context?
3. How would the calculation change if \( T \) was given in seconds instead of in terms of \( 2\pi \)?
4. Can you derive the formula for \( T \) from the principles of simple harmonic motion?
5. What units should \( k \) have to be consistent with the given formula?

**Tip:** When dealing with formulas involving square roots, always ensure you isolate the square root term before squaring both sides to simplify the equation.

Discover how to solve for the spring constant (k) using the formula for the period of oscillation, T = 2π √(m/k), with T=3.70 and m=67. This step-by-step guide covers essential concepts in algebra and simple harmonic motion, making it perfect for high school physics students (Grades 10-12).

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