Passage IX) For SHM to take place force acting on the body should be proportional to -x or F = -kx. If A be the amplitude then energy of oscillation is `1/2`k`A^(2)`
Force acting on a block is F = ( − 4 x + 8 ) .
Here F is in Newton and x the position of block on x-axis in meter If energy of osciallation is 18 J, between what points the block will oscillate?

Options

A. between x=0 and x=4m
B. between x=-1m and x=5m
C. between x=-2m and x=6m
D. between x=1m and x=3m

Answer

To solve the problem step by step, we will analyze the force acting on the block and relate it to the energy of oscillation in Simple Harmonic Motion (SHM).

### Step 1: Identify the force equation
The force acting on the block is given by:
\[ F = -4x + 8 \]

### Step 2: Rearrange the force equation
We can rearrange the force equation to match the standard form of SHM:
\[ F = -kx \]
To do this, we can factor out -4 from the force equation:
\[ F = -4(x - 2) \]
This indicates that the equilibrium position (mean position) is at \( x = 2 \) meters.

### Step 3: Find the spring constant \( k \)
From the rearranged force equation, we can see that:
\[ k = 4 \, \text{N/m} \]

### Step 4: Use the energy formula
The energy of oscillation in SHM is given by:
\[ E = \frac{1}{2} k A^2 \]
We know that the energy \( E \) is 18 J. Substituting the known values:
\[ 18 = \frac{1}{2} \times 4 \times A^2 \]

### Step 5: Solve for amplitude \( A \)
Now, we can solve for \( A^2 \):
\[ 18 = 2A^2 \]
\[ A^2 = \frac{18}{2} = 9 \]
\[ A = 3 \, \text{m} \]

### Step 6: Determine the oscillation limits
The block will oscillate between the extreme positions, which can be calculated as:
- Lower extreme position: \( x_{min} = \text{mean position} - A = 2 - 3 = -1 \, \text{m} \)
- Upper extreme position: \( x_{max} = \text{mean position} + A = 2 + 3 = 5 \, \text{m} \)

### Conclusion
The block will oscillate between the points:
\[ x = -1 \, \text{m} \, \text{and} \, x = 5 \, \text{m} \]

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