A particle oscillates with S.H.M. according to the equation `x = 10 cos ( 2pit + (pi)/(4))`. Its acceleration at `t = 1.5 s` is

Options

A. `69.78 ms^(-2)`
B. ` 139.56ms^(-2)`
C. ` 279.12ms^(-2)`
D. 0

Answer

To find the acceleration of a particle oscillating in Simple Harmonic Motion (S.H.M.) described by the equation \( x = 10 \cos(2\pi t + \frac{\pi}{4}) \) at \( t = 1.5 \, \text{s} \), we will follow these steps:

### Step 1: Identify the parameters from the equation
The given equation is:
\[ x = 10 \cos(2\pi t + \frac{\pi}{4}) \]
From this equation, we can identify:
- Amplitude \( A = 10 \, \text{m} \)
- Angular frequency \( \omega = 2\pi \, \text{rad/s} \)

### Step 2: Calculate the displacement \( x \) at \( t = 1.5 \, \text{s} \)
Substituting \( t = 1.5 \) into the equation:
\[ x = 10 \cos(2\pi \cdot 1.5 + \frac{\pi}{4}) \]
Calculating the argument of the cosine:
\[ 2\pi \cdot 1.5 = 3\pi \]
Thus,
\[ x = 10 \cos(3\pi + \frac{\pi}{4}) = 10 \cos\left(3\pi + \frac{\pi}{4}\right) \]

### Step 3: Simplify the angle
The angle \( 3\pi + \frac{\pi}{4} \) can be simplified:
\[ 3\pi + \frac{\pi}{4} = \frac{12\pi}{4} + \frac{\pi}{4} = \frac{13\pi}{4} \]
Now, we can find the cosine:
\[ \cos\left(\frac{13\pi}{4}\right) = \cos\left(2\pi + \frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \]
Thus,
\[ x = 10 \cdot \left(-\frac{1}{\sqrt{2}}\right) = -\frac{10}{\sqrt{2}} \, \text{m} \]

### Step 4: Calculate the acceleration using \( a = -\omega^2 x \)
The acceleration \( a \) can be calculated using the formula:
\[ a = -\omega^2 x \]
Substituting the values:
\[ \omega^2 = (2\pi)^2 = 4\pi^2 \]
Now substituting \( x \):
\[ a = -4\pi^2 \left(-\frac{10}{\sqrt{2}}\right) = 4\pi^2 \cdot \frac{10}{\sqrt{2}} \]
Calculating:
\[ a = \frac{40\pi^2}{\sqrt{2}} \approx 279.14 \, \text{m/s}^2 \]

### Final Answer
The acceleration of the particle at \( t = 1.5 \, \text{s} \) is approximately:
\[ a \approx 279.14 \, \text{m/s}^2 \]