The amplitude and the periodic time of a S.H.M. are 5 cm and 6 sec respectively. At a distance of 2.5 cm away from the mean position, the phase will be

Options

A. `(pi)/(3)`
B. `(pi)/(4)`
C. `(pi)/(6)`
D. `(5pi)/(12)`

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Identify the given values
- Amplitude (A) = 5 cm
- Periodic time (T) = 6 seconds

### Step 2: Calculate the angular frequency (ω)
The angular frequency (ω) is given by the formula:
\[ \omega = \frac{2\pi}{T} \]
Substituting the value of T:
\[ \omega = \frac{2\pi}{6} = \frac{\pi}{3} \text{ rad/sec} \]

### Step 3: Write the equation of motion for S.H.M.
The equation of motion for simple harmonic motion can be expressed as:
\[ x(t) = A \sin(\omega t) \]
Substituting the values of A and ω:
\[ x(t) = 5 \sin\left(\frac{\pi}{3} t\right) \]

### Step 4: Set the position to 2.5 cm and solve for time (t)
We need to find the time when the particle is at a distance of 2.5 cm from the mean position:
\[ 2.5 = 5 \sin\left(\frac{\pi}{3} t\right) \]
Dividing both sides by 5:
\[ \sin\left(\frac{\pi}{3} t\right) = \frac{2.5}{5} = 0.5 \]

### Step 5: Find the angle corresponding to sin(θ) = 0.5
The angle whose sine is 0.5 is:
\[ \frac{\pi}{3} t = \frac{\pi}{6} \text{ (since } \sin\left(\frac{\pi}{6}\right) = 0.5\text{)} \]

### Step 6: Solve for time (t)
Now, we can solve for t:
\[ t = \frac{\frac{\pi}{6}}{\frac{\pi}{3}} = \frac{1}{2} \text{ seconds} \]

### Step 7: Calculate the phase at t = 1/2 seconds
The phase (φ) at this time can be calculated using:
\[ \text{Phase} = \omega t = \frac{\pi}{3} \cdot \frac{1}{2} = \frac{\pi}{6} \]

### Final Answer
The phase when the particle is at a distance of 2.5 cm from the mean position is:
\[ \frac{\pi}{6} \]

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