A mass of 1 kg suspended from a spring whose force constant is 400 `Nm^(-1)`, executes simple harmonic motion. When the energy of the oscillator is 2J, the maximum acceleration experienced by maas will be

Options

A. `2ms^(-2)`
B. `4ms^(-2)`
C. `40ms^(-2)`
D. `400ms^(-2)`

Answer

To find the maximum acceleration experienced by a mass executing simple harmonic motion, we can follow these steps:

### Step 1: Understand the Energy of the Oscillator
The total mechanical energy (E) of a mass-spring system in simple harmonic motion is given by the formula:
\[ E = \frac{1}{2} k a^2 \]
where:
- \( E \) is the total energy (in joules),
- \( k \) is the spring constant (in N/m),
- \( a \) is the amplitude of oscillation (in meters).

### Step 2: Substitute the Given Values
From the problem, we know:
- \( E = 2 \, \text{J} \)
- \( k = 400 \, \text{N/m} \)

Substituting these values into the energy equation:
\[ 2 = \frac{1}{2} \times 400 \times a^2 \]

### Step 3: Solve for Amplitude (a)
Rearranging the equation to solve for \( a^2 \):
\[ 2 = 200 a^2 \]
\[ a^2 = \frac{2}{200} = 0.01 \]
Taking the square root:
\[ a = \sqrt{0.01} = 0.1 \, \text{m} \]

### Step 4: Calculate Angular Frequency (ω)
The relationship between the spring constant \( k \), mass \( m \), and angular frequency \( \omega \) is given by:
\[ k = m \omega^2 \]
Given that the mass \( m = 1 \, \text{kg} \):
\[ 400 = 1 \cdot \omega^2 \]
Thus,
\[ \omega^2 = 400 \]

### Step 5: Calculate Maximum Acceleration (A)
The maximum acceleration \( A \) in simple harmonic motion is given by:
\[ A = \omega^2 a \]
Substituting the values we have:
\[ A = 400 \times 0.1 \]
\[ A = 40 \, \text{m/s}^2 \]

### Conclusion
The maximum acceleration experienced by the mass is \( 40 \, \text{m/s}^2 \).

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