For any S.H.M., amplitude is 6 cm . If instantaneous potential energy is half the total energy then distance of particle from its mean position is

Options

A. 3 cm
B. 4.2 cm
C. 5.8 cm
D. 6 cm

Explanation

To solve the problem, we will follow these steps:

**Step 1: Understand the given information.**
- Amplitude (A) = 6 cm
- Instantaneous potential energy (U) = 1/2 × Total energy (E)

**Step 2: Write the formulas for potential energy and total energy in Simple Harmonic Motion (SHM).**
- The potential energy (U) at a distance x from the mean position is given by:
\[ U = \frac{1}{2} k x^2 \]
- The total energy (E) in SHM is constant and is given by:
\[ E = \frac{1}{2} k A^2 \]

**Step 3: Set up the equation based on the information provided.**
Since the potential energy is half of the total energy, we can write:
\[ U = \frac{1}{2} E \]
Substituting the formulas for U and E:
\[ \frac{1}{2} k x^2 = \frac{1}{2} \left(\frac{1}{2} k A^2\right) \]
This simplifies to:
\[ k x^2 = \frac{1}{2} k A^2 \]

**Step 4: Cancel out the common terms.**
Since k is common on both sides, we can cancel it out (assuming k ≠ 0):
\[ x^2 = \frac{1}{2} A^2 \]

**Step 5: Substitute the value of amplitude (A).**
Given that A = 6 cm, we can substitute this into the equation:
\[ x^2 = \frac{1}{2} (6)^2 \]
\[ x^2 = \frac{1}{2} \times 36 = 18 \]

**Step 6: Solve for x.**
Taking the square root of both sides:
\[ x = \sqrt{18} = 3\sqrt{2} \text{ cm} \]

**Step 7: Approximate the value of x.**
To find a numerical value, we can approximate \(\sqrt{2} \approx 1.414\):
\[ x \approx 3 \times 1.414 \approx 4.242 \text{ cm} \]
Rounding this, we can say:
\[ x \approx 4.2 \text{ cm} \]

**Final Answer:**
The distance of the particle from its mean position is approximately **4.2 cm**.

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