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 of finding the time taken by a particle executing Simple Harmonic Motion (SHM) to move from its mean position to half its amplitude, we can follow these steps:

### Step 1: Understand the parameters of SHM
The general equation for SHM is given by:
\[ x = a \sin(\omega t) \]
where:
- \( x \) is the displacement from the mean position,
- \( a \) is the amplitude,
- \( \omega \) is the angular frequency,
- \( t \) is the time.

### 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 \( T = 12 \, \text{s} \), 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 want to find the time \( t \) when the particle is at half its amplitude, which means:
\[ x = \frac{a}{2} \]
Substituting this into the SHM equation gives:
\[ \frac{a}{2} = a \sin(\omega t) \]

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

### Step 5: Solve for \( \omega t \)
The sine function equals \( \frac{1}{2} \) at:
\[ \omega t = \frac{\pi}{6} \]
This is because \( \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \).

### Step 6: Substitute \( \omega \) to find \( t \)
Now substituting \( \omega = \frac{\pi}{6} \):
\[ \frac{\pi}{6} = \frac{\pi}{6} t \]

### Step 7: Solve for \( t \)
To isolate \( t \), we can divide both sides by \( \frac{\pi}{6} \):
\[ t = 1 \, \text{s} \]

### Conclusion
The time taken by the particle to go directly from its mean position to half its amplitude is \( t = 1 \, \text{s} \).