The velocity of a particle in simple harmonic motion at displacement y from mean position is

Options

A. `omegasqrt(a^(2)+y^(2)`
B. `omegasqrt(a^(2)-y^(2)`
C. `omegay`
D. `omega^(2)sqrt(a^(2)-y^(2)`

Answer

To find the velocity of a particle in simple harmonic motion (SHM) at a displacement \( y \) from the mean position, we can follow these steps:

### Step 1: Understand the Motion
In simple harmonic motion, the displacement \( y \) of a particle can be expressed as:
\[ y = A \sin(\omega t) \]
where:
- \( A \) is the amplitude of the motion,
- \( \omega \) is the angular frequency,
- \( t \) is time.

### Step 2: Differentiate to Find Velocity
The velocity \( v \) of the particle is the time derivative of the displacement \( y \):
\[ v = \frac{dy}{dt} = A \omega \cos(\omega t) \]

### Step 3: Relate \( \cos(\omega t) \) to Displacement \( y \)
To express \( v \) in terms of \( y \), we need to relate \( \cos(\omega t) \) to \( y \). From the sine function, we know:
\[ \sin(\omega t) = \frac{y}{A} \]
Using the Pythagorean identity, we can find \( \cos(\omega t) \):
\[ \cos^2(\omega t) + \sin^2(\omega t) = 1 \]
Substituting \( \sin(\omega t) \):
\[ \cos^2(\omega t) + \left(\frac{y}{A}\right)^2 = 1 \]
This simplifies to:
\[ \cos^2(\omega t) = 1 - \left(\frac{y}{A}\right)^2 \]

### Step 4: Solve for \( \cos(\omega t) \)
Taking the square root gives us:
\[ \cos(\omega t) = \sqrt{1 - \left(\frac{y}{A}\right)^2} \]

### Step 5: Substitute Back to Find Velocity
Now, substituting \( \cos(\omega t) \) back into the velocity equation:
\[ v = A \omega \cos(\omega t) = A \omega \sqrt{1 - \left(\frac{y}{A}\right)^2} \]
This can be rewritten as:
\[ v = \omega \sqrt{A^2 - y^2} \]

### Final Answer
Thus, the velocity of the particle at displacement \( y \) from the mean position is:
\[ v = \omega \sqrt{A^2 - y^2} \]

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