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 can follow these steps:

### Step 1: Identify the displacement equation
The displacement of the particle is given by:
\[ x(t) = 10 \cos(2\pi t + \frac{\pi}{4}) \]

### Step 2: Differentiate to find velocity
The velocity \( v(t) \) is the first derivative of the displacement \( x(t) \) with respect to time \( t \):
\[ v(t) = \frac{dx}{dt} = -10 \cdot 2\pi \sin(2\pi t + \frac{\pi}{4}) = -20\pi \sin(2\pi t + \frac{\pi}{4}) \]

### Step 3: Differentiate to find acceleration
The acceleration \( a(t) \) is the derivative of the velocity \( v(t) \):
\[ a(t) = \frac{dv}{dt} = -20\pi \cdot 2\pi \cos(2\pi t + \frac{\pi}{4}) = -40\pi^2 \cos(2\pi t + \frac{\pi}{4}) \]

### Step 4: Substitute \( t = 1.5 \, \text{s} \) into the acceleration equation
Now we substitute \( t = 1.5 \) into the acceleration equation:
\[ a(1.5) = -40\pi^2 \cos(2\pi(1.5) + \frac{\pi}{4}) \]
Calculating \( 2\pi(1.5) \):
\[ 2\pi(1.5) = 3\pi \]
Thus:
\[ a(1.5) = -40\pi^2 \cos(3\pi + \frac{\pi}{4}) \]

### Step 5: Simplify the cosine term
Using the cosine addition formula:
\[ \cos(3\pi + \frac{\pi}{4}) = \cos(3\pi) \cos(\frac{\pi}{4}) - \sin(3\pi) \sin(\frac{\pi}{4}) \]
Since \( \cos(3\pi) = -1 \) and \( \sin(3\pi) = 0 \):
\[ \cos(3\pi + \frac{\pi}{4}) = -1 \cdot \frac{1}{\sqrt{2}} - 0 \cdot \frac{1}{\sqrt{2}} = -\frac{1}{\sqrt{2}} \]

### Step 6: Substitute back to find acceleration
Now substituting back:
\[ a(1.5) = -40\pi^2 \left(-\frac{1}{\sqrt{2}}\right) = \frac{40\pi^2}{\sqrt{2}} \]

### Step 7: Calculate the numerical value
Calculating \( \frac{40\pi^2}{\sqrt{2}} \):
\[ \pi^2 \approx 9.87 \quad \text{(using } \pi \approx 3.14\text{)} \]
Thus:
\[ \frac{40 \times 9.87}{\sqrt{2}} \approx \frac{395}{1.414} \approx 279.12 \, \text{m/s}^2 \]

### Conclusion
The acceleration at \( t = 1.5 \, \text{s} \) is approximately \( 279.12 \, \text{m/s}^2 \).