A particle which is attached to a spring oscillates horizontally with simple harmonic motion with a frequency of `1//pi` Hz and total energy of 10J. If the maximum speed of the particle is `0.4ms^(-1)`, what is the force constant of the spring? What will be the maximum potential energy of the spring during the motion?

Answer

To solve the problem step by step, we will follow the given information and apply the relevant formulas.

### Step 1: Identify the given information
- Frequency (f) = \( \frac{1}{\pi} \) Hz
- Total energy (E) = 10 J
- Maximum speed (Vmax) = 0.4 m/s

### Step 2: Calculate the angular frequency (ω)
The angular frequency (ω) is related to the frequency (f) by the formula:
\[ \omega = 2\pi f \]
Substituting the value of f:
\[ \omega = 2\pi \left(\frac{1}{\pi}\right) = 2 \text{ rad/s} \]

### Step 3: Relate maximum speed to amplitude
The maximum speed (Vmax) in simple harmonic motion is given by:
\[ V_{max} = \omega A \]
Where A is the amplitude. Rearranging the formula to find the amplitude:
\[ A = \frac{V_{max}}{\omega} \]
Substituting the values:
\[ A = \frac{0.4}{2} = 0.2 \text{ m} \]

### Step 4: Use the total energy formula to find the spring constant (k)
The total energy (E) in simple harmonic motion is given by:
\[ E = \frac{1}{2} k A^2 \]
Rearranging the formula to find k:
\[ k = \frac{2E}{A^2} \]
Substituting the values of E and A:
\[ k = \frac{2 \times 10}{(0.2)^2} = \frac{20}{0.04} = 500 \text{ N/m} \]

### Step 5: Calculate the maximum potential energy
The maximum potential energy (PEmax) in the spring is equal to the total energy (E) when the kinetic energy is zero:
\[ PE_{max} = E = 10 \text{ J} \]

### Final Answers
- The force constant of the spring (k) is \( 500 \, \text{N/m} \).
- The maximum potential energy of the spring during the motion is \( 10 \, \text{J} \).

---