A particle executes S.H.M with time period 12 s. The time taken by the particle to go directly from its mean position to half its amplitude.

Options

A. 4 s
B. 3 s
C. 6 s
D. 1 s

Answer

To solve the problem, we need to determine the time taken by a particle executing Simple Harmonic Motion (SHM) to move from its mean position to half of its amplitude. Here’s the step-by-step solution:

### Step 1: Understand the equation of SHM
The standard equation for SHM is given by:
\[ y(t) = A \sin(\omega t + \phi) \]
Where:
- \( y(t) \) is the displacement from the mean position,
- \( A \) is the amplitude,
- \( \omega \) is the angular frequency,
- \( t \) is the time,
- \( \phi \) is the phase constant.

Since the particle starts at the mean position, the phase constant \( \phi = 0 \). Thus, the equation simplifies to:
\[ y(t) = A \sin(\omega t) \]

### Step 2: Determine the angular frequency
The angular frequency \( \omega \) is related to the time period \( T \) by the formula:
\[ \omega = \frac{2\pi}{T} \]
Given that the time period \( T = 12 \) seconds, we can calculate \( \omega \):
\[ \omega = \frac{2\pi}{12} = \frac{\pi}{6} \, \text{rad/s} \]

### Step 3: Set up the equation for half the amplitude
We need to find the time taken to reach half of the amplitude, which is \( \frac{A}{2} \). Setting \( y(t) = \frac{A}{2} \):
\[ \frac{A}{2} = A \sin(\omega t) \]

### Step 4: Simplify the equation
Dividing both sides by \( A \) (assuming \( A \neq 0 \)):
\[ \frac{1}{2} = \sin(\omega t) \]

### Step 5: Solve for \( \omega t \)
We know that:
\[ \sin(\frac{\pi}{6}) = \frac{1}{2} \]
Thus, we can equate:
\[ \omega t = \frac{\pi}{6} \]

### Step 6: Substitute \( \omega \) and solve for \( t \)
Substituting \( \omega = \frac{\pi}{6} \):
\[ \frac{\pi}{6} t = \frac{\pi}{6} \]

To find \( t \), we can cancel \( \frac{\pi}{6} \) from both sides:
\[ t = 1 \, \text{second} \]

### Final Answer
The time taken by the particle to go directly from its mean position to half its amplitude is **1 second**.

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