A particle executes SHM of period `1.2 s` and amplitude `8cm`. Find the time it takes to travel `3cm` from the positive extremity of its oscillation. `[cos^(-1)(5//8) = 0.9rad]`

Options

A. `0.28s`
B. `0.32 s`
C. `0.17 s`
D. `0.42 s`

Explanation

To solve the problem step by step, we will follow the principles of Simple Harmonic Motion (SHM).

### Step 1: Understand the parameters of SHM
Given:
- Period (T) = 1.2 s
- Amplitude (A) = 8 cm

### 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:
\[ \omega = \frac{2\pi}{1.2} \approx 5.24 \, \text{rad/s} \]

### Step 3: Determine the position of the particle
The particle starts from the positive extremity (x = A = 8 cm) and moves towards the mean position (x = 0). We need to find the time it takes to travel 3 cm from the positive extremity.

The position of the particle at any time t can be expressed as:
\[ x(t) = A \cos(\omega t) \]
We need to find the time when the particle is at:
\[ x = A - 3 \, \text{cm} = 8 \, \text{cm} - 3 \, \text{cm} = 5 \, \text{cm} \]

### Step 4: Set up the equation
Setting the position equation equal to 5 cm:
\[ 5 = 8 \cos(\omega t) \]
Rearranging gives:
\[ \cos(\omega t) = \frac{5}{8} \]

### Step 5: Solve for ωt
Using the inverse cosine function:
\[ \omega t = \cos^{-1}\left(\frac{5}{8}\right) \]
Given in the problem, \(\cos^{-1}\left(\frac{5}{8}\right) = 0.9 \, \text{rad}\).

### Step 6: Solve for time (t)
Now we can find t:
\[ t = \frac{\omega t}{\omega} = \frac{0.9}{\omega} \]
Substituting the value of ω:
\[ t = \frac{0.9}{5.24} \approx 0.171 \, \text{s} \]

### Step 7: Final answer
Thus, the time it takes for the particle to travel 3 cm from the positive extremity is approximately:
\[ t \approx 0.171 \, \text{s} \]