When the displacement of a particle in SHM from the mean position is 4 cm, the force acting on the particle is 6 N. Then the force acting on it when its displacement is 6 cm from the mean positio is

Options

A. 3 N
B. 16 N
C. 8 N
D. 9 N

Answer

To solve the problem step by step, we can use the relationship between force and displacement in Simple Harmonic Motion (SHM). The force acting on a particle in SHM is given by Hooke's Law, which states:

\[ F = -kx \]

where:
- \( F \) is the force,
- \( k \) is the spring constant (a measure of stiffness),
- \( x \) is the displacement from the mean position.

### Step 1: Determine the spring constant \( k \)

From the problem, we know that when the displacement \( x = 4 \) cm, the force \( F = 6 \) N. We can rearrange the formula to find \( k \):

\[ F = kx \]

Substituting the known values:

\[ 6 = k \times 4 \]

Now, solve for \( k \):

\[ k = \frac{6}{4} = \frac{3}{2} \, \text{N/cm} \]

### Step 2: Calculate the force when the displacement is 6 cm

Now, we need to find the force when the displacement \( x = 6 \) cm. Using the same formula:

\[ F = kx \]

Substituting \( k = \frac{3}{2} \) N/cm and \( x = 6 \) cm:

\[ F = \left(\frac{3}{2}\right) \times 6 \]

Calculating this gives:

\[ F = \frac{3 \times 6}{2} = \frac{18}{2} = 9 \, \text{N} \]

### Final Answer

The force acting on the particle when its displacement is 6 cm from the mean position is **9 N**.

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