A particle executing SHM according to the equation `x=5cos(2pit+(pi)/(4))` in SI units. The displacement and acceleration of the particle at t=1.5 s is

Options

A. `-3.0m,100ms^(-2)`
B. `+2.54m,200ms^(-2)`
C. `-3.54m,140ms^(-2)`
D. `+3.55m,120ms^(-2)`

Explanation

To solve the problem, we need to find the displacement and acceleration of a particle executing Simple Harmonic Motion (SHM) given by the equation:

\[ x = 5 \cos(2\pi t + \frac{\pi}{4}) \]

at \( t = 1.5 \) seconds.

### Step 1: Find the displacement at \( t = 1.5 \) seconds

1. Substitute \( t = 1.5 \) seconds into the displacement equation:
\[ x = 5 \cos(2\pi \cdot 1.5 + \frac{\pi}{4}) \]

2. Calculate \( 2\pi \cdot 1.5 \):
\[ 2\pi \cdot 1.5 = 3\pi \]

3. Now substitute this back into the equation:
\[ x = 5 \cos(3\pi + \frac{\pi}{4}) \]

4. Simplify \( 3\pi + \frac{\pi}{4} \):
\[ 3\pi + \frac{\pi}{4} = \frac{12\pi}{4} + \frac{\pi}{4} = \frac{13\pi}{4} \]

5. Now calculate \( \cos\left(\frac{13\pi}{4}\right) \):
- The angle \( \frac{13\pi}{4} \) can be simplified by subtracting \( 3\pi \) (or \( \frac{12\pi}{4} \)):
\[ \frac{13\pi}{4} - 3\pi = \frac{13\pi}{4} - \frac{12\pi}{4} = \frac{\pi}{4} \]
- Thus, \( \cos\left(\frac{13\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \).

6. Substitute back into the displacement equation:
\[ x = 5 \cdot \frac{1}{\sqrt{2}} = \frac{5}{\sqrt{2}} \approx 3.54 \text{ meters} \]
Since \( 3\pi \) is in the third quadrant, \( \cos\left(\frac{13\pi}{4}\right) \) is negative:
\[ x \approx -3.54 \text{ meters} \]

### Step 2: Find the acceleration

1. The acceleration in SHM is given by:
\[ a = -\omega^2 x \]
where \( \omega = 2\pi \) rad/s.

2. Calculate \( \omega^2 \):
\[ \omega^2 = (2\pi)^2 = 4\pi^2 \]

3. Substitute \( x \) into the acceleration formula:
\[ a = -4\pi^2 \left(-3.54\right) = 4\pi^2 \cdot 3.54 \]

4. Calculate \( 4\pi^2 \):
\[ 4\pi^2 \approx 4 \cdot (3.14)^2 \approx 4 \cdot 9.86 \approx 39.44 \]

5. Finally, calculate the acceleration:
\[ a \approx 39.44 \cdot 3.54 \approx 140 \text{ m/s}^2 \]

### Summary of Results
- Displacement at \( t = 1.5 \) seconds: \( x \approx -3.54 \text{ meters} \)
- Acceleration at \( t = 1.5 \) seconds: \( a \approx 140 \text{ m/s}^2 \)