A mass of 4 kg suspended from a spring of spring constant `800 Nm^(-1)` executes simple harmonic oscillations. If the total energy of the oscillator is 4 J, the maximum acceleration (in `ms^(-2)`) of the mass is

Options

A. 5
B. 15
C. 45
D. 20

Explanation

To solve the problem, we need to find the maximum acceleration of a mass executing simple harmonic motion (SHM). Here are the steps to arrive at the solution:

### Step 1: Identify the given values
- Mass (m) = 4 kg
- Spring constant (k) = 800 N/m
- Total energy (E) = 4 J

### Step 2: Use the formula for total energy in SHM
The total energy (E) of a mass-spring system in simple harmonic motion is given by the formula:
\[ E = \frac{1}{2} k A^2 \]
where \( A \) is the amplitude.

### Step 3: Rearrange the formula to find the amplitude (A)
From the total energy formula, we can rearrange it to solve for \( A \):
\[ A^2 = \frac{2E}{k} \]
Substituting the known values:
\[ A^2 = \frac{2 \times 4 \, \text{J}}{800 \, \text{N/m}} \]
\[ A^2 = \frac{8}{800} = 0.01 \]
Taking the square root:
\[ A = \sqrt{0.01} = 0.1 \, \text{m} \]

### Step 4: Calculate the angular frequency (ω)
The angular frequency \( \omega \) is given by:
\[ \omega = \sqrt{\frac{k}{m}} \]
Substituting the values:
\[ \omega = \sqrt{\frac{800 \, \text{N/m}}{4 \, \text{kg}}} \]
\[ \omega = \sqrt{200} \]

### Step 5: Calculate the maximum acceleration (a_max)
The maximum acceleration \( a_{\text{max}} \) in SHM can be calculated using the formula:
\[ a_{\text{max}} = \omega^2 A \]
Substituting the values we have:
\[ a_{\text{max}} = (200) \times (0.1) \]
\[ a_{\text{max}} = 20 \, \text{m/s}^2 \]

### Final Answer
Thus, the maximum acceleration of the mass is:
\[ \boxed{20 \, \text{m/s}^2} \]