A particle executes simple harmonic motion with an amplitude of 4 cm . At the mean position the velocity of the particle is 10 cm/ s . The distance of the particle from the mean position when its speed becomes 5 cm/s is

Options

A. `sqrt3 cm`
B. `sqrt5 cm`
C. `2sqrt3 cm`
D. `2sqrt5 cm`

Answer

To solve the problem step by step, we will use the principles of simple harmonic motion (SHM).

### Step 1: Understand the given parameters
- Amplitude (A) = 4 cm
- Maximum velocity (v_max) at the mean position = 10 cm/s
- We need to find the distance from the mean position when the velocity (v) = 5 cm/s.

### Step 2: Write the equations for SHM
The position \( y \) of a particle in SHM can be expressed as:
\[ y = A \sin(\omega t) \]
where \( A \) is the amplitude.

The velocity \( v \) of the particle in SHM can be expressed as:
\[ v = A \omega \cos(\omega t) \]

### Step 3: Relate maximum velocity to angular frequency
At the mean position, the velocity is maximum:
\[ v_{\text{max}} = A \omega \]
Given that \( v_{\text{max}} = 10 \) cm/s and \( A = 4 \) cm, we can find \( \omega \):
\[ 10 = 4 \omega \]
\[ \omega = \frac{10}{4} = 2.5 \, \text{rad/s} \]

### Step 4: Use the velocity equation to find the position
We know that when the velocity \( v = 5 \) cm/s, we can substitute this into the velocity equation:
\[ v = A \omega \cos(\omega t) \]
Substituting the known values:
\[ 5 = 4 \cdot 2.5 \cos(\omega t) \]
\[ 5 = 10 \cos(\omega t) \]
\[ \cos(\omega t) = \frac{5}{10} = \frac{1}{2} \]

### Step 5: Use the relationship between position and velocity
Using the identity for SHM:
\[ \frac{y^2}{A^2} + \frac{v^2}{v_{\text{max}}^2} = 1 \]
Substituting the values we have:
\[ \frac{y^2}{4^2} + \frac{5^2}{10^2} = 1 \]
\[ \frac{y^2}{16} + \frac{25}{100} = 1 \]
\[ \frac{y^2}{16} + \frac{1}{4} = 1 \]

### Step 6: Solve for \( y^2 \)
Convert \( \frac{1}{4} \) to a fraction with a denominator of 16:
\[ \frac{y^2}{16} + \frac{4}{16} = 1 \]
\[ \frac{y^2 + 4}{16} = 1 \]
Multiply both sides by 16:
\[ y^2 + 4 = 16 \]
\[ y^2 = 16 - 4 \]
\[ y^2 = 12 \]

### Step 7: Find \( y \)
Taking the square root of both sides:
\[ y = \sqrt{12} = 2\sqrt{3} \]
Since the particle can be on either side of the mean position:
\[ y = \pm 2\sqrt{3} \, \text{cm} \]

### Final Answer
The distance of the particle from the mean position when its speed becomes 5 cm/s is \( \pm 2\sqrt{3} \) cm.

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