A particle executes SHM of period 1.2s and amplitude 8 cm. Find the time it takes to travel 13 cm from the positively extremily of its oscillation `cos^(-1)0.625=51^(@)`

Options

A. 0.27s
B. 0.12s
C. 0.21s
D. 0.17s

Explanation

To solve the problem, we need to find the time it takes for a particle executing simple harmonic motion (SHM) to travel 13 cm from the positive extremity of its oscillation. The given parameters are:

- Period (T) = 1.2 s
- Amplitude (A) = 8 cm

### Step-by-Step Solution:

1. **Understanding the Position in SHM**:
The particle's position \( y \) in SHM can be described by the equation:
\[ y = A \cos(\omega t) \]
where \( \omega \) is the angular frequency.

2. **Calculating Angular Frequency**:
The angular frequency \( \omega \) is given by:
\[ \omega = \frac{2\pi}{T} \]
Substituting the period:
\[ \omega = \frac{2\pi}{1.2} \approx 5.24 \, \text{rad/s} \]

3. **Finding the Position**:
The particle starts at the positive extremity (maximum displacement, which is equal to the amplitude, 8 cm). If it travels 13 cm from the positive extremity, its new position \( y \) is:
\[ y = A - 13 = 8 - 13 = -5 \, \text{cm} \]

4. **Setting Up the Equation**:
Now, we set up the equation:
\[ -5 = 8 \cos(\omega t) \]
Rearranging gives:
\[ \cos(\omega t) = \frac{-5}{8} \]

5. **Finding the Angle**:
To find \( \omega t \), we take the inverse cosine:
\[ \omega t = \cos^{-1}\left(-\frac{5}{8}\right) \]
Using the provided hint:
\[ \cos^{-1}(0.625) = 51^\circ \]
Since \( \cos(180^\circ - \theta) = -\cos(\theta) \):
\[ \omega t = 180^\circ - 51^\circ = 129^\circ \]

6. **Converting Degrees to Radians**:
Convert \( 129^\circ \) to radians:
\[ 129^\circ \times \frac{\pi}{180} \approx 2.25 \, \text{radians} \]

7. **Calculating Time**:
Now, we can solve for \( t \):
\[ t = \frac{\omega t}{\omega} = \frac{2.25}{5.24} \approx 0.43 \, \text{s} \]

### Final Answer:
The time it takes for the particle to travel 13 cm from the positive extremity is approximately **0.43 seconds**.