A body is in simple harmonic motion with time period T = 0.5 s and amplitude A = 1 cm. Find the average velocity in the interval in which it moves from equilibrium position to half of its amplitude.

Options

A. `16 cm//s`
B. `6 cm//s`
C. `4 cm//s`
D. `12 cm//s`

Answer

To find the average velocity of a body in simple harmonic motion (SHM) as it moves from the equilibrium position to half of its amplitude, we will follow these steps:

### Step 1: Understand the parameters
- The time period \( T \) is given as \( 0.5 \) seconds.
- The amplitude \( A \) is given as \( 1 \) cm.
- The equilibrium position corresponds to \( x = 0 \).
- Half of the amplitude is \( \frac{A}{2} = \frac{1 \text{ cm}}{2} = 0.5 \text{ cm} \).

### Step 2: Write the equation of motion
The displacement \( x \) in SHM can be expressed as:
\[ x(t) = A \sin(\omega t) \]
where \( \omega \) is the angular frequency given by:
\[ \omega = \frac{2\pi}{T} \]
Substituting \( T = 0.5 \) seconds:
\[ \omega = \frac{2\pi}{0.5} = 4\pi \text{ rad/s} \]

### Step 3: Set up the equation for half amplitude
We want to find the time \( t \) when the displacement \( x(t) \) is \( \frac{A}{2} \):
\[ \frac{A}{2} = A \sin(\omega t) \]
Substituting \( A = 1 \text{ cm} \):
\[ 0.5 = 1 \sin(4\pi t) \]
This simplifies to:
\[ \sin(4\pi t) = 0.5 \]

### Step 4: Solve for \( t \)
The value of \( t \) can be found using the inverse sine function:
\[ 4\pi t = \sin^{-1}(0.5) \]
Since \( \sin^{-1}(0.5) = \frac{\pi}{6} \):
\[ 4\pi t = \frac{\pi}{6} \]
Dividing both sides by \( 4\pi \):
\[ t = \frac{1}{24} \text{ seconds} \]

### Step 5: Calculate the average velocity
The average velocity \( V_{\text{avg}} \) is defined as the total displacement divided by the total time taken:
\[ V_{\text{avg}} = \frac{\text{Displacement}}{\text{Time}} = \frac{\frac{A}{2}}{t} \]
Substituting the values:
\[ V_{\text{avg}} = \frac{0.5 \text{ cm}}{\frac{1}{24} \text{ s}} = 0.5 \times 24 = 12 \text{ cm/s} \]

### Conclusion
The average velocity of the body as it moves from the equilibrium position to half of its amplitude is:
\[ \boxed{12 \text{ cm/s}} \]