Questions: A mass of 2.00 kg connected to a spring of spring constant 500.0 N/m undergoes simple harmonic motion with an amplitude of 30.0 cm. What is the period of oscillation?

Solution Steps

The period of oscillation for a mass-spring system undergoing simple harmonic motion is given by the formula:

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

where \( T \) is the period, \( m \) is the mass, and \( k \) is the spring constant.

Given:

• Mass, \( m = 2.00 \, \text{kg} \)
• Spring constant, \( k = 500.0 \, \text{N/m} \)

Substitute these values into the formula:

\[ T = 2\pi \sqrt{\frac{2.00}{500.0}} \]

First, calculate the value inside the square root:

\[ \frac{2.00}{500.0} = 0.004 \]

Now, calculate the square root:

\[ \sqrt{0.004} = 0.06325 \]

Finally, calculate the period:

\[ T = 2\pi \times 0.06325 \approx 0.3979 \]

Final Answer